2024.04 别急记录

1. 餐巾计划问题

建图跑费用流即可：

$(S, 1, inf, p)$ ;
$\forall i \in [1, N], (i, i + N, r_{i}, 0)$ ;
$\forall i \in [1, N], (S, i + N, r_{i}, 0)$ ;
$\forall i \in [1, N], (i, T, r_{i}, 0)$ ;
$\forall i \in [1, N), (i, i + 1, inf, 0)$ ;
$\forall i \in [1, N - n], (i + N, i + n, inf, f)$ ;
$\forall i \in [1, N - m], (i + N, i + m, inf, s)$ ;

解释一下 234：

考虑把餐巾作为流，费用作为费用。则我们需要保证最大流为 $\sum r$ 。所以每个点要拆成两部分，表示干净毛巾和脏毛巾，再用脏毛巾的点流向洗过后的干净毛巾点。

点击查看代码

 //P1251
#include <bits/stdc++.h>
using namespace std; typedef long long ll;
void solve();int main(){ solve(); return 0; }
 
const int N = 4e3 + 10, M = 4e4 + 10;
const ll inf = 1e10;
int n, r[N], p, u, f, v, s;
ll mc, ln[M], cs[M], ds[N];
int hd[N], eg[M], nx[M], tot = 1, nw[N], vs[N];
 
void adg(int u, int v, ll w, ll c){
	eg[++tot] = v;
	ln[tot] = w;
	cs[tot] = c;
	nx[tot] = hd[u];
	hd[u] = tot;
}
void add(int u, int v, ll w, ll c){
//	printf("%d %d %lld %lld\n", u, v, w, c);
	adg(u, v, w, c);
	adg(v, u, 0, -c);
}
bool spfa(int s, int t){
	queue<int> q;
	memset(ds, 0x3f, sizeof(ds));
	memcpy(nw, hd, sizeof(hd));
	q.push(s);
	ds[s] = 0;
	vs[s] = 1;
	while(!q.empty()){
		int x = q.front();
		q.pop();
		vs[x] = 0;
		for(int i = hd[x]; i; i = nx[i]){
			int y = eg[i];
			ll z = ln[i], c = cs[i];
			if(z && ds[y] > ds[x] + c){
				ds[y] = ds[x] + c;
				if(!vs[y]){
					vs[y] = 1;
					q.push(y);
				}
			}
		}
	}
	return ds[t] != 0x3f3f3f3f3f3f3f3f;
}
ll dfs(int x, int t, ll fl){
	if(x == t){
		return fl;
	}
	ll rs = fl;
	vs[x] = 1;
	for(int i = nw[x]; i && rs; i = nx[i]){
		int y = eg[i];
		ll z = ln[i], c = cs[i];
		nw[x] = i;
		if(!vs[y] && z && ds[y] == ds[x] + c){
			ll k = dfs(y, t, min(rs, z));
			if(!k){
				ds[y] = 0;
			}
			ln[i] -= k;
			ln[i^1] += k;
			rs -= k;
			mc += k * c;
		}
	}
	vs[x] = 0;
	return fl - rs;
}
 
void solve(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++ i){
		scanf("%d", &r[i]);
	}
	scanf("%d%d%d%d%d", &p, &u, &f, &v, &s);
	int st = n + n + 1, ed = n + n + 2;
	for(int i = 1; i <= n; ++ i){
		add(i, i+n, r[i], 0);
		add(st, i+n, r[i], 0);
		add(i, ed, r[i], 0);
		if(i < n){
			add(i, i+1, inf, 0);
		}
		if(i + v <= n){
			add(i+n, i+v, inf, s);
		}
		if(i + u <= n){
			add(i+n, i+u, inf, f);
		}
	}
	add(st, 1, inf, p);
	while(spfa(st, ed)){
		while(dfs(st, ed, inf*2));
	}
	printf("%lld\n", mc);
}

2. 国家集训队 - happiness

考虑答案=总和-最小割。

对于点 $i$ ，连接 $(S, i), (i, T)$ 表示文和理，割掉表示不选。
对于额外贡献 $(i, j, p)$ ，若为文则新建点 $x$ 连 $(S, x, p), (x, i, inf), (x, j, inf)$ ，否则连 $(i, x, inf), (j, x, inf), (x, T, p)$ 。

点击查看代码

 //P1646
#include <bits/stdc++.h>
using namespace std;
 
const int N = 5e4 + 10, inf = 100000, M = 3e5 + 10;
int n, m, hd[N], eg[M], ln[M], nx[M], tot = 1, now[N], dep[N];
 
void adg(int u, int v, int w){
    eg[++tot] = v;
    ln[tot] = w;
    nx[tot] = hd[u];
    hd[u] = tot;
}
void add(int u, int v, int w){
    adg(u, v, w);
    adg(v, u, 0);
}
bool bfs(int s, int t){
    queue<int> q;
    q.push(s);
    memcpy(now, hd, sizeof(hd));
    memset(dep, 0, sizeof(dep));
    dep[s] = 1;
    while(!q.empty()){
        int x = q.front();
        q.pop();
        for(int i = hd[x]; i; i = nx[i]){
            int y = eg[i], z = ln[i];
            if(z && !dep[y]){
                dep[y] = dep[x] + 1;
                q.push(y);
                if(y == t){
                    return 1;
                }
            }
        }
    }
    return 0;
}
int dfs(int x, int t, int fl){
    if(x == t){
        return fl;
    } else {
        int rs = fl;
        for(int i = now[x]; i && rs; i = nx[i]){
            int y = eg[i], z = ln[i];
            now[x] = i;
            if(z && dep[y] == dep[x] + 1){
                int k = dfs(y, t, min(rs, z));
                if(!k){
                    dep[y] = 0;
                }
                ln[i] -= k;
                ln[i^1] += k;
                rs -= k;
            }
        }
        return fl - rs;
    }
}
 
int main(){
    #define cg(x, y) ((x - 1) * m + y)
    scanf("%d%d", &n, &m);
    int sm = 0;
    int st = cg(n, m) + 1, ed = cg(n, m) + 2, a, cnt = cg(n, m) + 2;
    for(int i = 1; i <= n; ++ i){
        for(int j = 1; j <= m; ++ j){
            scanf("%d", &a);
            sm += a;
            add(st, cg(i, j), a);
        }
    }
    for(int i = 1; i <= n; ++ i){
        for(int j = 1; j <= m; ++ j){
            scanf("%d", &a);
            sm += a;
            add(cg(i, j), ed, a);
        }
    }
    for(int i = 1; i < n; ++ i){
        for(int j = 1; j <= m; ++ j){
            scanf("%d", &a);
            sm += a;
            ++ cnt;
            add(st, cnt, a);
            add(cnt, cg(i, j), inf);
            add(cnt, cg(i+1, j), inf);
        }
    }
    for(int i = 1; i < n; ++ i){
        for(int j = 1; j <= m; ++ j){
            scanf("%d", &a);
            sm += a;
            ++ cnt;
            add(cnt, ed, a);
            add(cg(i, j), cnt, inf);
            add(cg(i+1, j), cnt, inf);
        }
    }
    for(int i = 1; i <= n; ++ i){
        for(int j = 1; j < m; ++ j){
            scanf("%d", &a);
            sm += a;
            ++ cnt;
            add(st, cnt, a);
            add(cnt, cg(i, j), inf);
            add(cnt, cg(i, j+1), inf);
        }
    }
    for(int i = 1; i <= n; ++ i){
        for(int j = 1; j < m; ++ j){
            scanf("%d", &a);
            sm += a;
            ++ cnt;
            add(cnt, ed, a);
            add(cg(i, j), cnt, inf);
            add(cg(i, j+1), cnt, inf);
        }
    }
    int mf = 0, tmp;
    while(bfs(st, ed)){
        while(tmp = dfs(st, ed, inf)) mf += tmp; 
    }
    printf("%d\n", sm - mf);
    return 0;
}

3. qoj1427 - Flip

考虑 $\sum k = 2 * 10^{5}$ 的话不同 $k$ 只有根号个，而相同 $k$ 之间可以直接算。

然后 $k \geq 450$ 的也只有根号个，可以根号分治。

两个部分分别都可以根号做。

点击查看代码

 //qoj1427
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
 
const ll P = 998244353;
const int N = 2e5 + 10;
int n, m, c, a[N];
ll fac[N], inv[N], pw[N], ivy, ans[N];
vector<pair<int, int> > g[N], h[N];
 
ll qp(ll x, ll y){
    ll ans = 1;
    while(y){
        if(y & 1){
            ans = ans * x % P;
        }
        x = x * x % P;
        y >>= 1;
    }
    return ans;
}
ll C(int x, int y){
    if(x < y) return 0;
    return fac[x] * inv[y] % P * inv[x-y] % P;
}
 
int main(){
    fac[0] = pw[0] = 1;
    for(int i = 1; i < N; ++ i){
        fac[i] = fac[i-1] * i % P;
        pw[i] = pw[i-1] * 2 % P;
    }
    inv[N-1] = qp(fac[N-1], P-2);
    for(int i = N - 2; i >= 0; -- i){
        inv[i] = inv[i+1] * (i+1) % P;
    }
    scanf("%d%d", &n, &m);
    ivy = qp(pw[n+n], P-2);
    for(int cs = 1; cs <= m; ++ cs){
        scanf("%d", &c);
        for(int i = 1; i <= c; ++ i){
            scanf("%d", &a[i]);
        }
        if(a[c] < n + n){
            ans[cs] = (ans[cs] + C(a[c]-c, n-c) * pw[n+n-a[c]]) % P;
        }
        g[c].push_back(make_pair(max(n, a[c]+1), cs));
        int ls = n - 1;
        if(c > 450){
            for(int i = 1; i <= c; ++ i){
                if(a[i] > ls){
                    for(int k = ls+1; k < a[i]; ++ k){
                        ans[cs] = (ans[cs] + C(k-i, n-1) * pw[n+n-k]) % P;
                    }
                    ls = a[i];
                }
            }
            h[c+1].push_back(make_pair(ls+1, cs));
            h[c+1].push_back(make_pair(n+n, -cs));
        } else {
            for(int i = 1; i <= c; ++ i){
                if(a[i] > ls){
                    h[i].push_back(make_pair(ls+1, cs));
                    h[i].push_back(make_pair(a[i], -cs));
                    ls = a[i];
                }
            }
            h[c+1].push_back(make_pair(ls+1, cs));
            h[c+1].push_back(make_pair(n+n, -cs));
        }
    }
    for(int i = 1; i <= n; ++ i){
        if(g[i].size()){
            sort(g[i].begin(), g[i].end());
            reverse(g[i].begin(), g[i].end());
            int nw = n + n;
            ll rs = 0;
            for(auto j : g[i]){
                while(nw > j.first){
                    -- nw;
                    rs = (rs + C(nw-1-i, n-1-i) * pw[n+n-nw]) % P;
                }
                ans[j.second] = (ans[j.second] + rs) % P;
            }
        }
    }
    for(int i = 1; i <= n + 1; ++ i){
        if(h[i].size()){
            sort(h[i].begin(), h[i].end());
            reverse(h[i].begin(), h[i].end());
            int nw = n + n;
            ll rs = 0;
            for(auto j : h[i]){
                while(nw > j.first){
                    -- nw;
                    rs = (rs + C(nw-i, n-1) * pw[n+n-nw]) % P;
                }
                if(j.second > 0){
                    ans[j.second] = (ans[j.second] + rs) % P;
                } else {
                    ans[-j.second] = (ans[-j.second] - rs + P) % P;
                }
            }
        }
    }
    for(int i = 1; i <= m; ++ i){
        printf("%lld\n", 2 * ans[i] * ivy % P);
    }
    return 0;
}

4. USACO 2024 OPEN - Cowreography G

考虑第一个串中 '1' 的位置为 $x_{1}, x_{2}, . . ., x_{k}$ ，第二个串中为 $y_{1}, y_{2}, . . ., y_{k}$ ，则我们可以对这 $k$ 对进行匹配，一对 $x, y$ 的代价为 $⌈ \frac{| y - x |}{k} ⌉$ 考虑后面的去匹配前面的。则从左往右遍历，维护两个集合表示 '1' 在上面的串还是下面的，那么对于位置 $i$ ，优先找的位置为与 $i$ 模 $k$ 同余的，其次找模 $k$ 余数比 $i$ 模 $k$ 大 $1$ 的，以此类推。

点击查看代码

 //P10280
#include <bits/stdc++.h>
using namespace std;
 
const int N = 1e6 + 10;
int n, k;
char s[N], t[N];
long long ans;
set<pair<int, int> > st[2];
 
int main(){
    scanf("%d%d", &n, &k);
    scanf("%s%s", s+1, t+1);
    for(int i = 1; i <= n; ++ i){
        if(s[i] != t[i]){
            int x = s[i] - '0';
            if(st[!x].empty()){
                st[x].insert(make_pair(i%k, i));
            } else {
                auto it = st[!x].lower_bound(make_pair(i % k, 0));
                if(it == st[!x].end()){
                    it = st[!x].begin();
                }
                ans += (i - (*it).second - 1) / k + 1;
                st[!x].erase(it);
            }
        }
    }
    printf("%lld\n", ans);
    return 0;
}

5. CCO2019 - Sirtet

首先 dfs 出所有连通块。对于一个连通块 $X$ ，可以定义一个变量 $d_{X}$ 表示下落后连通块 $X$ 最低点在哪一行。预处理出连通块中每个点 $(i, j)$ 比它所在连通块最低点高了 $t_{(i, j)}$ 行，则每个点下落后的位置为 $x - t_{(i, j)} + 1$ 。

对于两个连通块 $A, B$ ，若 $(p, y) \in A, (q, y) \in B, p < q$ ，则我们可以得出 $d_{A} - t_{(p, y)} + 1 < d_{B} - t_{(q, y)} + 1$ ，是一个差分约束的形式。所以可以列出所有这样的限制后建图跑一遍差分约束即可。

由于建 $0$ 边跑最短路差分约束所求的解是 $\leq 0$ 的最大值，所以对于每个数 $+ n$ 即为最终的位置。

不卡 spfa 啊。

点击查看代码

 //P5532
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
 
const int N = 5e6 + 10;
const int dx[4] = {0, -1, 1, 0}, dy[4] = {-1, 0, 0, 1};
int plv[N], pcl[N], vs[N], ds[N], n, m, cs;
vector<pair<int, int> > g[N];
char pl[N], ans[N];
 
void dfs(int x, int y, int l, int c){
    char (&s)[n+3][m+3] = decltype(s)(pl);
    int (&cl)[n+3][m+3] = decltype(cl)(pcl);
    int (&lv)[n+3][m+3] = decltype(lv)(plv);
    lv[x][y] = l;
    cl[x][y] = c;
    for(int i = 0; i < 4; ++ i){
        int xx = x + dx[i], yy = y + dy[i];
        if(s[xx][yy] == '#' && !lv[xx][yy]){
            dfs(xx, yy, l-dx[i], c);
        }
    }
}
void spfa(int st){
    queue<int> q;
    memset(vs, 0, sizeof(vs));
    memset(ds, 0x3f, sizeof(ds));
    vs[st] = 1, ds[st] = 0, q.push(st);
    while(!q.empty()){
        int x = q.front();
        q.pop();
        vs[x] = 0;
        for(auto i : g[x]){
            if(ds[i.first] > ds[x] + i.second){
                ds[i.first] = ds[x] + i.second;
                if(!vs[i.first]){
                    q.push(i.first);
                    vs[i.first] = 1;
                }
            }
        }
    }
}
 
int main(){
    scanf("%d%d", &n, &m);
    char (&s)[n+3][m+3] = decltype(s)(pl);
    char (&t)[n+3][m+3] = decltype(t)(ans);
    int (&cl)[n+3][m+3] = decltype(cl)(pcl);
    int (&lv)[n+3][m+3] = decltype(lv)(plv);
    for(int i = 1; i <= n; ++ i){
        scanf("%s", s[i] + 1);
    }
    for(int i = n; i >= 1; -- i){
        for(int j = 1; j <= m; ++ j){
            t[i][j] = '.';
            if(!lv[i][j] && s[i][j] == '#'){
                dfs(i, j, 1, ++ cs);
            }
        }
    }
    for(int j = 1; j <= m; ++ j){
        int ls = 0;
        for(int i = n; i >= 1; -- i){
            if(s[i][j] != '#'){
                continue;
            }
            if(ls && cl[i][j] != cl[ls][j]){
                g[cl[ls][j]].emplace_back(cl[i][j], lv[i][j]-lv[ls][j]-1);
            }
            ls = i;
        }
    }
    for(int i = 1; i <= cs; ++ i){
        g[cs+1].emplace_back(i, 0);
    }
    spfa(cs + 1);
    for(int i = 1; i <= n; ++ i){
        for(int j = 1; j <= m; ++ j){
            if(s[i][j] == '#'){
                t[ds[cl[i][j]]+n-lv[i][j]+1][j] = '#';
            }
        }
    }
    for(int i = 1; i <= n; ++ i){
        printf("%s\n", t[i] + 1);
    }
    return 0;
}

6. LuoguP3600 - 随机数生成器

水黑逆天 $O (n x \log n)$ 做法！

考虑设 $f_{i}$ 表示 $max min \leq T$ 的方案数，则此时每个数可以分为 $> T, \leq T$ 两部分，合法当且仅当每个区间内存在数 $\leq T$ 。可以 $O (n^{2})$ 做。然后使用线段树优化 dp 即可做到 $O (n x \log n)$ 。

点击查看代码

 //P3600
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
 
const int N = 2010;
const ll P = 666623333;
int n, x, q, mr[N];
ll f[N][N], g[N];
 
ll qp(ll x, ll y){
    ll ans = 1;
    while(y){
        if(y & 1){
            ans = ans * x % P;
        }
        x = x * x % P;
        y >>= 1;
    }
    return ans;
}
 
ll t[N*4], tag[N*4];
void psd(int p){
    t[p<<1] = t[p<<1] * tag[p] % P;
    t[p<<1|1] = t[p<<1|1] * tag[p] % P;
    tag[p<<1] = tag[p<<1] * tag[p] % P;
    tag[p<<1|1] = tag[p<<1|1] * tag[p] % P;
    tag[p] = 1;
}
void mul(int p, int l, int r, int ql, int qr, ll v){
    if(ql <= l && r <= qr){
        t[p] = t[p] * v % P;
        tag[p] = tag[p] * v % P;
    } else if(r < ql || qr < l){
        return;
    } else {
        int mid = l + r >> 1;
        psd(p);
        mul(p<<1, l, mid, ql, qr, v);
        mul(p<<1|1, mid+1, r, ql, qr, v);
        t[p] = (t[p<<1] + t[p<<1|1]) % P;
    }
}
void add(int p, int l, int r, int x, ll v){
    if(l == r){
        t[p] = (t[p] + v) % P;
    } else {
        int mid = l + r >> 1;
        psd(p);
        if(x <= mid){
            add(p<<1, l, mid, x, v);
        } else {
            add(p<<1|1, mid+1, r, x, v);
        }
        t[p] = (t[p<<1] + t[p<<1|1]) % P;
    }
}
ll qry(int p, int l, int r, int ql, int qr){
    if(ql <= l && r <= qr){
        return t[p];
    } else if(r < ql || qr < l){
        return 0;
    } else {
        int mid = l + r >> 1;
        psd(p);
        return (qry(p<<1, l, mid, ql, qr) +
               qry(p<<1|1, mid+1, r, ql, qr)) % P;
    }
}
 
int main(){
    scanf("%d%d%d", &n, &x, &q);
    for(int i = 1; i <= q; ++ i){
        int l, r;
        scanf("%d%d", &l, &r);
        mr[r] = max(mr[r], l);
    }
    for(int T = 1; T <= x; ++ T){
        mul(1, 0, n, 0, n, 0);
        add(1, 0, n, 0, 1);
        for(int i = 1; i <= n; ++ i){
            ll val = t[1] * T % P;
            t[1] = t[1] * (x-T) % P;
            tag[1] = tag[1] * (x-T) % P;
            mul(1, 0, n, 0, mr[i]-1, 0);
            add(1, 0, n, i, val);
        }
        g[T] = t[1];
    }
    ll ans = 0;
    for(int i = n; i >= 1; -- i){
        g[i] = (g[i] - g[i-1] + P) % P;
        ans += g[i] * i % P;
        ans %= P;
    }
    printf("%lld\n", ans * qp(qp(x, n), P-2) % P);
    return 0;
}

7. AGC066A - Adjacent Difference

设 $f (x)$ 表示与 $x$ 最接近的为 $d$ 奇数倍倍数的数； $g (x)$ 表示与 $x$ 最接近的为 $d$ 偶数倍倍数的数，则容易发现有 $| f (x) - g (x) | = d$ 且二者一个 $\leq x$ 一个 $\geq x$ 。

考虑对矩阵黑白染色，相邻位置不同色，有两种末方案：

黑色位置取 $f (x)$ ，白色位置取 $g (x)$ ；
黑色位置取 $g (x)$ ，白色位置取 $f (x)$ ；

容易发现二者的代价和为 $d n^{2}$ ，故存在一个代价 $\leq \frac{1}{2} d n^{2}$ 。

点击查看代码

 //AT_agc066_a
#include <bits/stdc++.h>
using namespace std; typedef long long ll;
void solve();int main(){ solve(); return 0; }
 
const int N = 510;
int n, d, a[N][N], b[N][N], c[N][N], sb, sc;
 
void solve(){
	scanf("%d%d", &n, &d);
	for(int i = 1; i <= n; ++ i){
		for(int j = 1; j <= n; ++ j){
			scanf("%d", &a[i][j]);
			a[i][j] += 1000;
			if(((i + j) & 1) ^ ((a[i][j] / d) & 1)){
				b[i][j] = a[i][j] - a[i][j] % d;
				c[i][j] = b[i][j] + d;
				sb += a[i][j] - b[i][j];
				sc += c[i][j] - a[i][j];
			} else {
				c[i][j] = a[i][j] - a[i][j] % d;
				b[i][j] = c[i][j] + d;
				sc += a[i][j] - c[i][j];
				sb += b[i][j] - a[i][j];
			}
		}
	}
	for(int i = 1; i <= n; ++ i){
		for(int j = 1; j <= n; ++ j){
			printf("%d ", (sb < sc ? b[i][j] : c[i][j]) - 1000);
		}
		puts("");
	}
}

8. AGC066B - Decreasing Digit Sums

考虑对于形如 $5^{k}$ 的数，比如 $125 \to 250 \to 500 \to 1000$ ，看似直接满足条件了，但是 $5^{k}$ 的数位和并不是递增的，于是我们可以把 $5^{k}, 5^{k - 1}, . . ., 5$ 拼起来，如 $. . .3125625125255$ ，可以发现这个数多次 $\times 2$ 会变成：

$. . .3, 125, 625, 125, 255$ ；
$. . .6, 251, 250, 250, 510$ ；
$. . .12, 502, 500, 501, 020$ ；
$. . .25, 005, 001, 002, 040$ ；
$. . .50, 010, 002, 004, 080$ 。

假设最高位是 $5^{k}$ ，那么每次操作后数位和相当于减少 $f (5^{k - x})$ ，增加 $f (2^{x})$ （ $x \in [0, n]$ ）。当 $k$ 尽可能大的时候更有可能成功。打表可得 $k = 160$ 时可行且总长度 $< 10000$ 。

点击查看代码

 //AT_agc066_b
#include <bits/stdc++.h>
using namespace std; typedef long long ll;
void solve();int main(){ solve(); return 0; }
 
const int N = 10010;
int s[N], t[N];
int n, len = 1, tot = 1;
 
void solve(){
	cin >> n;
	t[1] = s[1] = 5;
	for(int i = 2; i <= 160; ++ i){
		for(int j = 1; j <= len; ++ j){
			t[j] = t[j] * 5;
			t[j] = t[j] + t[j-1] / 10;
			t[j-1] %= 10;
		}
		while(t[len] >= 10){
			t[len+1] = t[len] / 10;
			t[len] %= 10;
			++ len;
		}
		for(int j = 1; j <= len; ++ j){
			s[++tot] = t[j];
		}
	}
	for(int i = tot; i >= 1; -- i){
		printf("%d", s[i]);
	}
	puts("");
}

9. AGC066C - Delete AAB or BAA

一个结论：

若一个串能够被删空，当且仅当它满足以下两个条件，或它能拆分成若干个满足以下两个条件的子串：

A 的数量为 B 的两倍。

首尾中有一个是 B。

对于本身满足两个条件的串 $S, | S | = 3 k$ ，考虑归纳证明：

若 $k = 1$ ，AAB，BAA 满足条件；
找到一段长度 $\geq 2$ 的极长 A 段，它的两侧各为空或者一个 B，容易发现至少有一个 B 不在两侧，删掉这个 B 以及相邻的两个 A 变成了 $3 (k - 1)$ 的子问题且依旧满足条件 2。

而对于任意一个能够被删空的串，考虑从空串往中间插入，容易发现肯定也满足条件。

所以可以进行 dp。设 $f_{i}$ 表示 $[1, i]$ 至少要保留几个字符，转移为 $f_{i} = f_{i - 1} + 1$ 以及当 $(j, i]$ 满足条件时 $f_{i} = f_{j}$ 。

设 $s u m_{i} = \sum_{j \leq i} [S_{j} =^{'} A^{'}] 1 - [S_{j} =^{'} B^{'}] 2,$ ，则第一个条件可以写成 $s u m_{i} = s u m_{j}$ 。

点击查看代码

 //AT_agc066_c
#include <bits/stdc++.h>
using namespace std; typedef long long ll;
void solve();int main(){ solve(); return 0; }
 
const int N = 5e6 + 10;
int T, n, f[N], g[N], h[N];
char s[N];
 
void solve(){
	scanf("%d", &T);
	while(T--){
		scanf("%s", s + 1);
		n = strlen(s + 1);
		int sum = N / 2;
		for(int i = sum - n*2; i <= sum + n*2; ++ i){
			g[i] = h[i] = 0x3f3f3f3f;
		}
		for(int i = 0; i <= n; ++ i){
			if(i){
				f[i] = f[i-1] + 1;
				if(s[i] == 'A'){
					++ sum;
					f[i] = min(f[i], g[sum]);
				} else {
					-- sum;
					-- sum;
					f[i] = min(f[i], g[sum]);
					f[i] = min(f[i], h[sum]);
				}
			}
			if(s[i+1] == 'B'){
				g[sum] = min(g[sum], f[i]);
			} else {
				h[sum] = min(h[sum], f[i]);
			}
		}
		printf("%d\n", (n - f[n]) / 3);
		for(int i = 0; i <= n; ++ i){
			f[i] = 0;
		}
	}
}

10. CF526F - Pudding Monsters

考虑从左往右枚举右端点 $r$ ，动态维护一棵线段树，位置 $l$ 的值为 $t_{l} = max [l, r] - min [l, r] + l$ ，那么若一个 $[l, r]$ 可行当且仅当 $t_{l} = r$ 。观察到 $t_{l} \geq r$ ，所以只用求线段树上 $[1, r]$ 区间内最小值个数即可。

新增右端点 $r$ 时，可以维护两个单调栈，在维护的同时更新 $max, min$ 。一棵支持区间加+区间最值个数查询的线段树即可维护。

点击查看代码

 //CF526F
#include <bits/stdc++.h>
using namespace std; typedef long long ll;
void solve();int main(){ solve(); return 0; }
 
const int N = 3e5 + 10;
int n, a[N], mx[N], mn[N], tx, tn;
ll ans = 0;
 
struct node{
	int tag, mn, cnt;
} t[N*4];
 
void psd(int p){
	t[p<<1].tag += t[p].tag;
	t[p<<1].mn += t[p].tag;
	t[p<<1|1].tag += t[p].tag;
	t[p<<1|1].mn += t[p].tag;
	t[p].tag = 0; 
}
 
void build(int p, int l, int r){
	t[p].cnt = r - l + 1;
	if(l == r){
		return;
	}
	int mid = l + r >> 1;
	build(p<<1, l, mid);
	build(p<<1|1, mid+1, r);
}
void add(int p, int l, int r, int ql, int qr, int x){
	if(qr < l || r < ql){
		return;	
	} else if(ql <= l && r <= qr){
		t[p].mn += x;
		t[p].tag += x;
	} else {
		int mid = l + r >> 1;
		psd(p);
		add(p<<1, l, mid, ql, qr, x);
		add(p<<1|1, mid+1, r, ql, qr, x);
		t[p].mn = min(t[p<<1].mn, t[p<<1|1].mn);
		if(t[p<<1].mn < t[p<<1|1].mn){
			t[p].cnt = t[p<<1].cnt;
		} else if(t[p<<1].mn > t[p<<1|1].mn){
			t[p].cnt = t[p<<1|1].cnt;
		} else {
			t[p].cnt = t[p<<1].cnt + t[p<<1|1].cnt;
		}
	}
}
pair<int, int> qry(int p, int l, int r, int ql, int qr){
	if(qr < l || r < ql){
		return make_pair(1e9, 0);
	} else if(ql <= l && r <= qr){
		return make_pair(t[p].mn, t[p].cnt);
	} else {
		int mid = l + r >> 1;
		psd(p);
		pair<int, int> x = qry(p<<1, l, mid, ql, qr);
		pair<int, int> y = qry(p<<1|1, mid+1, r, ql, qr);
		if(x.first < y.first){
			return x;
		} else if(x.first > y.first){
			return y;
		} else {
			return make_pair(x.first, x.second + y.second);
		}
	}
}
 
void solve(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++ i){
		int x, y;
		scanf("%d%d", &x, &y);
		a[x] = y;
	}
	build(1, 1, n);
	for(int i = 1; i <= n; ++ i){
		add(1, 1, n, i, i, i);
		while(tx && a[mx[tx]] < a[i]){
			add(1, 1, n, mx[tx-1]+1, mx[tx], -a[mx[tx]]);
			-- tx;
		}
		add(1, 1, n, mx[tx]+1, i, a[i]);
		mx[++tx] = i;
		while(tn && a[mn[tn]] > a[i]){
			add(1, 1, n, mn[tn-1]+1, mn[tn], a[mn[tn]]);
			-- tn;
		}
		add(1, 1, n, mn[tn]+1, i, -a[i]);
		mn[++tn] = i;
		ans += qry(1, 1, n, 1, i).second;
	}
	printf("%lld\n", ans);
}

11. NOI2023 - 贸易

一条路径 $(x, y)$ 可以拆分成 $(x, l c a), (l c a, y)$ 两部分。第一部分直接是树边权值和；第二部分可能是 $l c a$ 先向上若干格，走一条非树边到达子数中。可以首先处理这些非树边目标点 $d i s$ 然后跑一遍 dij 解决。

点击查看代码

 //P9481
//You can get an NOI2023 Ag by solving this problem!
//You can get an NOI2023 Au by 509pts.
//You can get 5pts by O(3^n) 后面忘了
//509+5=514
//then 进入集训队！！！
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
 
const int N = 262150;
const ll P = 998244353;
int n, m, a[N], dep[N], vis[N];
vector<pair<int, int> > g[N];
vector<int> eg[N], son[N];
ll sum[N], dis[N], ans, fa[N][20];
struct edge{
    int u, v, w;
} e[N];
 
int main(){
    scanf("%d%d", &n, &m);
    n = (1 << n) - 1;
    dep[1] = 1;
    son[1].push_back(1);
    for(int i = 2; i <= n; ++ i){
        son[i].push_back(i);
        scanf("%d", &a[i]);
        dep[i] = dep[i>>1] + 1;
        for(int j = 1; j <= dep[i]; ++ j){
            fa[i][j] = fa[i][j-1] + a[i>>(j-1)];
            sum[i>>j] = (sum[i>>j] + fa[i][j]) % P;
            son[i>>j].push_back(i);
        }
    }
    for(int i = 1; i <= m; ++ i){
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        e[i] = { u, v, w };
        g[u].emplace_back(v, w);
        int x = v;
        while(x != u){
            eg[x].push_back(i);
            x >>= 1;
        }
    }
    for(int rt = 1; rt <= n / 2; ++ rt){
        priority_queue<pair<ll, int> > q;
        for(int j : son[rt]){
            dis[j] = 1e18;
            vis[j] = 0;
        }
        dis[rt] = 0;
        q.push(make_pair(-dis[rt], rt));
        while(!q.empty()){
            int x = q.top().second;
            q.pop();
            if(vis[x] == 1){
                continue;
            }
            vis[x] = 1;
            if(x == rt){
                for(int i : eg[x]){
                    int y = e[i].v;
                    ll z = e[i].w;
                    z += fa[x][dep[x] - dep[e[i].u]];
                    if(dis[y] > dis[x] + z){
                        dis[y] = dis[x] + z;
                        q.push(make_pair(-dis[y], y));
                    }
                }
            }
            for(auto i : g[x]){
                int y = i.first;
                ll z = i.second;
                if(dis[y] > dis[x] + z){
                    dis[y] = dis[x] + z;
                    q.push(make_pair(-dis[y], y));
                }
            }
            if((x>>1) >= rt && dis[x>>1] > dis[x] + a[x]){
                dis[x>>1] = dis[x] + a[x];
                q.push(make_pair(-dis[x>>1], x>>1));
            }
        }
        ll le = 0, cle = 1;
        for(int j : son[rt<<1]){
            if(dis[j] > 1e16){
                continue;
            }
            ++ cle;
            le = (le + dis[j]) % P;
        }
        ans = (ans + cle * (sum[rt<<1|1] + a[rt<<1|1] * son[rt<<1|1].size() % P)) % P;
        ans = (ans + le * (son[rt<<1|1].size() + 1)) % P;
        cle = 1, le = 0;
        for(int j : son[rt<<1|1]){
            if(dis[j] > 1e16){
                continue;
            }
            ++ cle;
            le = (le + dis[j]) % P;
        }
        ans = (ans + cle * (sum[rt<<1] + a[rt<<1] * son[rt<<1].size() % P)) % P;
        ans = (ans + le * (son[rt<<1].size() + 1)) % P;
    }
    printf("%lld\n", ans);
    return 0;
}

12. THUPC 2024 初赛 - 多折较差验证

首先预处理 $o k_{l, r}$ 表示 $[l, r]$ 可以被折空。显然有 $O (n^{3})$ 的转移。考虑 $f_{l, r}$ 的最优折叠点一定是中点两侧离中点最近的两个合法折叠位置。于是预处理出这两个位置暴力转移即可 $O (n^{2})$ 。

点击查看代码

 //P9964
#include <bits/stdc++.h>
using namespace std;
 
const int N = 5010;
int n, f[N][N][2], ok[N][N], g[N][N][2];
char s[N];
 
int main(){
    scanf("%d%s", &n, s+1);
    for(int len = 1; len <= n; len += 2){
        for(int l = 1; l + len - 1 <= n; ++ l){
            int r = l + len - 1;
            if(l == r){
                ok[l][r] = 1;
            } else {
                ok[l][r] = (s[l] != s[r]) & ok[l+1][r-1];
            }
            if(ok[l][r]){
                g[l][r][0] = g[l][r][1] = (l+r) >> 1;
            }
        }
    }
    memset(f, 0x3f, sizeof(f));
    for(int i = 1; i <= n; ++ i){
        for(int j = i; j <= n; ++ j){
            if(g[i][j][0] == 0){
                g[i][j][0] = g[i][j-1][0];
            }
        }
    }
    for(int i = n; i >= 1; -- i){
        for(int j = i; j >= 1; -- j){
            if(g[j][i][1] == 0){
                g[j][i][1] = g[j+1][i][1];
            }
        }
    }
    for(int len = 1; len <= n; ++ len){
        for(int l = 1; l + len - 1 <= n; ++ l){
            int r = l + len - 1;
            if(l == r){
                f[l][r][0] = 1;
                f[l][r][1] = 0;
            } else {
                int x = g[l][r][0], y = g[l][r][1];
                if(f[x+1][r][0] + 1 < f[l][r][0]){
                    f[l][r][0] = f[x+1][r][0] + 1;
                    f[l][r][1] = f[x+1][r][1] + r + l - x - x;
                } else if(f[x+1][r][0] + 1 == f[l][r][0]){
                    f[l][r][1] = min(f[l][r][1], f[x+1][r][1] + r + l - x - x);
                }
                if(f[l][y-1][0] + 1 < f[l][r][0]){
                    f[l][r][0] = f[l][y-1][0] + 1;
                    f[l][r][1] = f[l][y-1][1] + y + y - r - l;
                } else if(f[l][y-1][0] + 1 == f[l][r][0]){
                    f[l][r][1] = min(f[l][r][1], f[l][y-1][1] + y + y - r - l);
                }
            }
        }
    }
    printf("%d %d\n", f[1][n][0], f[1][n][1]);
    return 0;
}

13. AGC035C - Skolem XOR Tree

显然 $n = 2^{k}$ 时无解。

其他情况如图构造：

点击查看代码

 //AT_agc035_c
#include <bits/stdc++.h>
using namespace std;
 
const int N = 1e5 + 10;
int n;
 
int main(){
    scanf("%d", &n);
    int p = 1;
    while(p < n){
        p *= 2;
    }
    if(p == n){
        puts("No");
    } else {
        puts("Yes");
        printf("%d %d\n", 1+n, 3);
        for(int i = 2; i <= n - 1; i += 2){
            printf("%d %d\n", i, 1);
            printf("%d %d\n", i, i+1);
            printf("%d %d\n", i+n+1, 1);
            printf("%d %d\n", i+n+1, i+n);
        }
        if(n % 2 == 0){
            printf("%d %d\n", n, n-2);
            printf("%d %d\n", n+n, n+(n^(n-2)^1));
        }
    }
    return 0;
}

14. CF1874E - Jellyfish and Hack

首先可以设 $f_{p, q}$ 表示长度为 $p$ ， $fun ()$ 值为 $q$ 的排列个数，容易写出转移方程：

$f_{p, q} = \sum_{i = 1}^{p} (\binom{p - 1}{i - 1}) \sum_{j = 0}^{q - p} f_{i - 1, j} f_{p - i, q - p - j}$ 。

设 $F_{p} (x) = \sum_{i = 0}^{\infin} f_{p, q} x^{q}$ ，则有 $F_{p} (x) = x^{p} \sum_{i = 1}^{p} (\binom{p - 1}{i - 1}) F_{i - 1} (x) F_{p - i} (x)$ 。

设 $m = \frac{n (n + 1)}{2} + 1$ 。

观察到若 $2 q > p (p + 1)$ ，则 $f_{p, q} = 0$ 。所以 $F_{n}$ 至多 $m - 1$ 次。可以使用上述转移式计算出 $F_{n} (1), F_{n} (2), . . ., F_{n} (m)$ 的值。

接下来可以倒着用拉格朗日插值，已知点值求出表达式。设 $x_{i} = i, y_{i} = F_{n} (i)$ 。

$y = \sum_{i = 1}^{m} y_{i} \prod_{j \neq i} \frac{1}{x_{i} - x_{j}} [\prod_{j} (x - x_{j}) \frac{1}{x - x_{i}}]$

枚举 $i$ ，中括号以外的直接预处理；以内的可以先预处理出 $\prod_{j} (x - x_{j})$ 再使用竖式除法除掉 $x - x_{i}$ 。

总复杂度 $O (m^{2}) = O (n^{4})$ 。

点击查看代码

 //CF1874E
#include <bits/stdc++.h>
using namespace std; typedef long long ll;
void solve();int main(){ solve(); return 0; }
 
const int N = 210;
const ll P = 1e9 + 7;
ll y[N*N], C[N][N], F[N];
ll fac[N*N], f[N*N], p[N*N], q[N*N];
int n, lim, m;
 
ll qp(ll x, ll y){
	ll ans = 1;
	while(y){
		if(y & 1){
			ans = ans * x % P;
		}
		x = x * x % P;
		y >>= 1;
	}
	return ans;
}
 
void solve(){
	cin >> n >> lim;
	m = n * (n + 1) / 2 + 1;
	for(int i = 0; i <= n; ++ i){
		C[i][0] = C[i][i] = 1;
		for(int j = 1; j < i; ++ j){
			C[i][j] = (C[i-1][j] + C[i-1][j-1]) % P;
		}
	}
	fac[0] = 1;
	for(int i = 1; i <= m; ++ i){
		fac[i] = fac[i-1] * i % P;
		F[0] = 1;
		ll pw = 1;
		for(int k = 1; k <= n; ++ k){
			pw = pw * i % P;
			F[k] = 0;
			for(int j = 1; j <= k; ++ j){
				F[k] = (F[k] + C[k-1][j-1] * F[j-1] % P * F[k-j]) % P;
			}
			F[k] = F[k] * pw % P;
		}
		y[i] = F[n];
	}
	p[0] = 1;
	for(int i = 1; i <= m; ++ i){
		for(int j = 1; j <= i; ++ j){
			q[j] = p[j-1];
			p[j-1] = p[j-1] * (P - i) % P;
		}
		for(int j = 0; j <= i; ++ j){
			p[j] = (p[j] + q[j]) % P;
		}
	}
	for(int i = 1; i <= m; ++ i){
		ll val = y[i] * qp(fac[i-1], P-2) % P * qp(fac[m-i], P-2) % P;
		if((m - i) & 1){
			val = P - val;
		}
		for(int j = 0; j <= m; ++ j){
			q[j] = p[j];
		}
		for(int j = m - 1; j >= 0; -- j){
			ll r = q[j+1];
			f[j] = (f[j] + r * val) % P;
			q[j] = (q[j] + r * i) % P;
		}
	}
	ll ans = 0;
	for(int i = lim; i < m; ++ i){
		ans = (ans + f[i]) % P;
	}
	cout << ans << '\n';
}

15. AGC030C - Coloring Torus

这你都不会？

首先 $k \leq 500$ 是平凡的。令 $a_{x, i} = x$ 即可。

对于 $k > 500$ 的情况，可以对于前 $500$ 个数，把一个数 $i$ 填到 $a_{1, i}, a_{2, i + 1}, . . ., a_{500, i - 1}$ 中；若 $i + 500 \leq k$ ，则把奇数行的 $i$ 换成 $500 + i$ 。

点击查看代码

 //AT_agc030_c
#include <bits/stdc++.h>
using namespace std; typedef long long ll;
void solve();int main(){ solve(); return 0; }
 
const int N = 510;
int k, a[N][N];
 
void solve(){
//	freopen(".out", "w", stdout);
	scanf("%d", &k);
	if(k <= 500){
		printf("%d\n", k);
		for(int i = 1; i <= k; ++ i){
			for(int j = 1; j <= k; ++ j){
				printf("%d ", j);
			}
			puts("");
		}
	} else {
		puts("500");
		for(int i = 1; i <= 500; ++ i){
			for(int j = 1, p = i; j <= 500; ++ j){
				a[j][p] = i;
				if(i + 500 <= k && j % 2 == 0){
					a[j][p] = i + 500;
				}
				++ p;
				if(p > 500){
					p -= 500;
				}
			}
		}
		for(int i = 1; i <= 500; ++ i){
			for(int j = 1; j <= 500; ++ j){
				printf("%d ", a[i][j]);
			}
			puts("");
		}
	}
}

16. NOI2020 - 超现实树

离正解就差一点/ll。

结论：若一棵树中存在一个节点的两个子结点均有子结点，则这棵树无效。

结论成立，因为这些数不会对任何其中一个儿子为空/其中一个儿子为叶子的树产生贡献。

有了这个结论然后使用四叉 trie 维护其中一个儿子为空/其中一个儿子为叶子共四种情况。然后从叶子往根 dp 一下即可。

点击查看代码

 //P6776
#include <bits/stdc++.h>
using namespace std;
 
const int N = 2e6 + 10;
int T, m, n;
vector<pair<int, int> > g[N];
int ch[N][4], tot = 1, ok[N];
 
int sc(pair<int, int> x){
    return (x.first ? 1 : 0) + (x.second ? 1 : 0);
}
void dfs(int i, int p, int q){
    if(g[i][q].first && g[i][q].second){
        if(!sc(g[i][g[i][q].first])){
            if(!ch[p][2]) ch[p][2] = ++ tot;
            dfs(i, ch[p][2], g[i][q].second);
        }
        if(!sc(g[i][g[i][q].second])){
            if(!ch[p][3]) ch[p][3] = ++ tot;
            dfs(i, ch[p][3], g[i][q].first);
        }
    } else if(g[i][q].first){
        if(!ch[p][0]) ch[p][0] = ++ tot;
        dfs(i, ch[p][0], g[i][q].first);
    } else if(g[i][q].second){
        if(!ch[p][1]) ch[p][1] = ++ tot;
        dfs(i, ch[p][1], g[i][q].second);
    } else {
        ok[p] = 1;
    }
}
 
int main(){
    scanf("%d", &T);
    while(T--){
        scanf("%d", &m);
        for(int i = 1; i <= m; ++ i){
            scanf("%d", &n);
            g[i].resize(n + 1);
            for(int j = 1; j <= n; ++ j){
                scanf("%d%d", &g[i][j].first, &g[i][j].second);
            }
            bool flg = 1;
            for(int j = 1; j <= n; ++ j){
                if(sc(g[i][g[i][j].first]) && sc(g[i][g[i][j].second])){
                    flg = 0;
                    break;
                }
            }
            if(flg){
                dfs(i, 1, 1);
            }
        }
        for(int i = tot; i >= 1; -- i){
            if(ok[ch[i][0]] && ok[ch[i][1]] && ok[ch[i][2]] && ok[ch[i][3]]){
                ok[i] = 1;
            }
        }
        puts(ok[1] ? "Almost Complete" : "No");
        for(int i = 0; i <= tot; ++ i){
            ch[i][0] = ch[i][1] = ch[i][2] = ch[i][3] = ok[i] = 0;
        }
        tot = 1;
    }
    return 0;
}

17. NOI2020 - 制作菜品

显然当 $m \geq n - 1$ 时可以直接贪心：取最小的和最大的中的一部分，转化为 $(n - 1, m - 1)$ 的问题。

当 $m = n - 2$ 时，考虑对于每个物品的两部分连一条边建图，则图不连通，即可以拆分成两个小问题，小问题中分别 $m = n - 1$ 。使用背包求一下划分方案即可。

点击查看代码

 //P6775
#include <bits/stdc++.h>
using namespace std;
 
const int N = 510;
int T, n, m, k, ok[N];
pair<int, int> d[N], p[N];
bitset<N*N*20> b[N];
 
void solve(int cnt, int l, int r){
    int tp = l;
    for(int i = 1; i <= cnt; ++ i){
        sort(d + tp, d + r + 1);
        if(d[tp].first > k){
            printf("%d %d", d[tp].second, k);
            d[tp].first -= k;
        } else if(d[tp].first == k){
            printf("%d %d", d[tp].second, k);
            d[tp].first = 0;
            ++ tp;
        } else {
            printf("%d %d ", d[tp].second, d[tp].first);
            printf("%d %d", d[r].second, k - d[tp].first);
            d[r].first -= (k - d[tp].first);
            d[tp].first = 0;
            ++ tp;
        }
        puts("");
    }
}
 
int main(){
    scanf("%d", &T);
    while(T--){
        scanf("%d%d%d", &n, &m, &k);
        for(int i = 1; i <= n; ++ i){
            scanf("%d", &d[i].first);
            d[i].second = i;
        }
        if(m >= n - 1){
            solve(m, 1, n);
        } else {
            int tmp = n * k + 3;
            b[0].reset();
            b[0].set(tmp);
            for(int i = 1; i <= n; ++ i){
                if(d[i].first >= k){
                    b[i] = b[i-1] | (b[i-1] << d[i].first - k);
                } else {
                    b[i] = b[i-1] | (b[i-1] >> k - d[i].first);
                }
            }
            if(b[n].test(tmp - k)){
                int flg = tmp - k;
                int l = 0, r = 0;
                for(int i = n; i >= 1; -- i){
                    if(b[i-1].test(flg)){
                        continue;
                    }
                    for(int j = 1; j <= n; ++ j){
                        if(!ok[i] && b[i-1].test(flg-d[i].first+k)){
                            ok[i] = 1;
                            flg -= d[i].first-k;
                            ++ r;
                        }
                    }
                }
                for(int i = 1; i <= n; ++ i){
                    if(ok[i] == 1){
                        p[++l] = d[i];
                    } else {
                        p[++r] = d[i];
                    }
                    ok[i] = 0;
                }
                memcpy(d, p, sizeof(p));
                solve(l-1, 1, l);
                solve(r-l-1, l+1, r);
            } else {
                puts("-1");
            }
        }
    }
    return 0;
}

18. LuoguP2633 - Count on a tree

使用主席树维护每个节点到根的值域线段树，接着使用类似树上差分的方法查询即可。

点击查看代码

 //P2633
#include <bits/stdc++.h>
using namespace std; typedef long long ll;
void solve();int main(){ solve(); return 0; }
 
const int N = 1e5 + 10;
int n, m, rg, a[N], b[N], anc[N][20], dep[N];
int t[N*20], ls[N*20], rs[N*20], rt[N], cnt = 1;
vector<int> g[N];
 
int add(int p, int l, int r, int x){
	int nw = ++ cnt;
	tie(t[nw], ls[nw], rs[nw]) = tie(t[p], ls[p], rs[p]);
	if(l == r){
		++ t[nw];
	} else {
		int mid = l + r >> 1;
		if(x <= mid){
			ls[nw] = add(ls[nw], l, mid, x);
		} else {
			rs[nw] = add(rs[nw], mid+1, r, x);
		}
		t[nw] = t[ls[nw]] + t[rs[nw]];
	}
	return nw;
}
int qry(int p, int q, int x, int y, int l, int r, int k){
	if(l == r){
		return l;
	} else {
		int mid = l + r >> 1;
		int val = t[ls[p]] + t[ls[q]] - t[ls[x]] - t[ls[y]];
		if(val >= k){
			return qry(ls[p], ls[q], ls[x], ls[y], l, mid, k);
		} else {
			return qry(rs[p], rs[q], rs[x], rs[y], mid+1, r, k-val);
		}
	}
}
 
void dfs(int x, int fa){
	anc[x][0] = fa;
	dep[x] = dep[fa] + 1;
	rt[x] = add(rt[fa], 1, rg, a[x]);
	for(int i = 1; i < 20; ++ i){
		anc[x][i] = anc[anc[x][i-1]][i-1];
	}
	for(int i : g[x]){
		if(i != fa){
			dfs(i, x);
		}
	}
}
int lca(int x, int y){
	if(dep[x] < dep[y]){
		swap(x, y);
	}
	for(int i = 19; i >= 0; -- i){
		if(dep[anc[x][i]] >= dep[y]){
			x = anc[x][i];
		}
	}
	if(x == y){
		return x;
	}
	for(int i = 19; i >= 0; -- i){
		if(anc[x][i] != anc[y][i]){
			x = anc[x][i];
			y = anc[y][i];
		}
	}
	return anc[x][0];
}
 
void solve(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; ++ i){
		scanf("%d", &a[i]);
		b[i] = a[i];
	}
	sort(b + 1, b + n + 1);
	rg = unique(b + 1, b + n + 1) - b - 1;
	for(int i = 1; i <= n; ++ i){
		a[i] = lower_bound(b + 1, b + rg + 1, a[i]) - b;
	}
	for(int i = 1; i < n; ++ i){
		int x, y;
		scanf("%d%d", &x, &y);
		g[x].push_back(y);
		g[y].push_back(x);
	}
	rt[0] = ++ cnt;
	dfs(1, 0);
	int ls = 0;
	while(m--){
		int u, v, k;
		scanf("%d%d%d", &u, &v, &k);
		u ^= ls;
		int p = lca(u, v);
		ls = b[qry(rt[u], rt[v], rt[p], rt[anc[p][0]], 1, rg, k)];
		printf("%d\n", ls);
	}
}

19. 国家集训队 - middle

考虑二分答案。让序列中的 $a_{i}$ 变为 $(- 1) [a_{i} < m i d] + (1) [a_{i} \geq m i d]$ ，则一个区间中位数 $\geq m i d$ 当且仅当 $s u m \geq 0$ 。

于是我们可以使用两棵主席树，分别维护当 $m i d = k$ 时序列前缀和/后缀和。使用 $r t_{k - 1}$ 推 $r t_{k}$ 只需区间加操作即可。

查询时则对于二分到的每个 $m i d$ ，查询左端点在 $[a, b]$ 之间最大后缀、右端点在 $[c, d]$ 之间最大前缀即可。

点击查看代码

 //P2839
#include <bits/stdc++.h>
using namespace std; typedef long long ll;
void solve();int main(){ solve(); return 0; }
 
const int N = 20010, M = 400;
int lrt[N], rrt[N], t[N*M], tag[N*M], ls[N*M], rs[N*M], cnt;
int n, a[N], q, la, pos[N];
pair<int, int> b[N];
 
int build(int l, int r, int op){
	int p = ++ cnt;
	if(l == r){
		t[p] = op ? l : n - l + 1;
	} else {
		int mid = l + r >> 1;
		ls[p] = build(l, mid, op);
		rs[p] = build(mid+1, r, op);
		t[p] = max(t[ls[p]], t[rs[p]]);
	}
	return p;
}
int nnd(int p){
	++ cnt;
	tie(t[cnt], tag[cnt], ls[cnt], rs[cnt]) = tie(t[p], tag[p], ls[p], rs[p]);
	return cnt;
}
void psd(int p){
	if(!tag[p]){
		return;
	}
	ls[p] = nnd(ls[p]);
	rs[p] = nnd(rs[p]);
	tag[ls[p]] += tag[p];
	tag[rs[p]] += tag[p];
	t[ls[p]] += tag[p];
	t[rs[p]] += tag[p];
	tag[p] = 0;
}
int add(int p, int l, int r, int ql, int qr){
	p = nnd(p);
	if(ql <= l && r <= qr){
		tag[p] -= 2;
		t[p] -= 2;
	} else {
		int mid = l + r >> 1;
		psd(p);
		if(ql <= mid){
			ls[p] = add(ls[p], l, mid, ql, qr);
		}
		if(mid < qr){
			rs[p] = add(rs[p], mid+1, r, ql, qr);
		}
		t[p] = max(t[ls[p]], t[rs[p]]);
	}
	return p;
}
int ask(int p, int l, int r, int ql, int qr){
	if(ql <= l && r <= qr){
		return t[p];
	} else {
		int mid = l + r >> 1;
		psd(p);
		int ans = -2e9;
		if(ql <= mid){
			ans = max(ans, ask(ls[p], l, mid, ql, qr));
		}
		if(mid < qr){
			ans = max(ans, ask(rs[p], mid+1, r, ql, qr));
		}
		return ans;
	}
}
 
bool chk(int a, int b, int c, int d, int k){
	-- k;
	int val = 0;
	val += ask(lrt[k], 1, n, c, d) - ask(lrt[k], 1, n, b, b);
	val += ask(rrt[k], 1, n, a, b) - ask(rrt[k], 1, n, b+1, b+1);
	return val >= 0;
}
 
void solve(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++ i){
		scanf("%d", &a[i]);
		b[i] = make_pair(a[i], i);
	}
	sort(b + 1, b + n + 1);
	for(int i = 1; i <= n; ++ i){
		a[i] = lower_bound(b + 1, b + n + 1, make_pair(a[i], i)) - b;
		pos[a[i]] = i;
	}
	lrt[0] = build(1, n, 1);
	rrt[0] = build(1, n, 0);
	for(int i = 1; i <= n; ++ i){
		lrt[i] = add(lrt[i-1], 1, n, pos[i], n);
		rrt[i] = add(rrt[i-1], 1, n, 1, pos[i]);
	}
	scanf("%d", &q);
	while(q--){
		int p[4];
		scanf("%d%d%d%d", &p[0], &p[1], &p[2], &p[3]);
		p[0] = (p[0] + la) % n;
		p[1] = (p[1] + la) % n;
		p[2] = (p[2] + la) % n;
		p[3] = (p[3] + la) % n;
		sort(p, p + 4);
		int l = 1, r = n;
		while(l < r){
			int mid = l + r + 1 >> 1;
			if(chk(p[0] + 1, p[1] + 1, p[2] + 1, p[3] + 1, mid)){
				l = mid;
			} else {
				r = mid - 1;
			}
		}
		la = b[l].first;
		printf("%d\n", la);
	}
}

20. HEOI/TJOI2016 - 字符串

我都会，这不降紫？

问题相当于求最长的 $[c, d]$ 中前缀使得这个前缀是 $[a, b]$ 中的子串。求出 $r k, h e i g h t$ 数组。

考虑二分答案。设目前二分到 $m i d$ ，则与后缀 $[c, n]$ 的 $l c p \geq m i d$ 的串的 $r k$ 为一个区间 $[x, y]$ ，可以使用二分+rmq 求出。

问题转化为 $r k_{[a, b]}$ 中是否有 $\in [x, y]$ 的数，是 IOI2018 狼人的套路，可以使用主席树做。

点击查看代码

 //P4094
#include <bits/stdc++.h>
using namespace std; typedef long long ll;
void solve();int main(){ solve(); return 0; }
 
const int N = 1e5 + 10, M = 30;
int n, q;
 
int sa[N], rk[N], cnt[N], pre[N], h[N];
char s[N];
void suf(){
	int m = 26;
	for(int i = 1; i <= n; ++ i){
		rk[i] = s[i] - 'a' + 1;
		++ cnt[rk[i]];
	}
	for(int i = 2; i <= m; ++ i){
		cnt[i] += cnt[i-1];
	}
	for(int i = n; i >= 1; -- i){
		sa[cnt[rk[i]]--] = i;
	}
	for(int k = 1; k <= n; k <<= 1){
		int tot = 0;
		for(int i = n - k + 1; i <= n; ++ i){
			pre[++tot] = i;
		}
		for(int i = 1; i <= n; ++ i){
			if(sa[i] > k){
				pre[++tot] = sa[i] - k;
			}
		}
		for(int i = 1; i <= m; ++ i){
			cnt[i] = 0;
		}
		for(int i = 1; i <= n; ++ i){
			++ cnt[rk[i]];
		}
		for(int i = 2; i <= m; ++ i){
			cnt[i] += cnt[i-1];
		}
		for(int i = n; i >= 1; -- i){
			sa[cnt[rk[pre[i]]]--] = pre[i];
			pre[i] = 0;
		}
		swap(rk, pre);
		tot = 1, rk[sa[1]] = 1;
		for(int i = 2; i <= n; ++ i){
			if(pre[sa[i]] == pre[sa[i-1]] && pre[sa[i]+k] == pre[sa[i-1]+k]){
				rk[sa[i]] = tot;
			} else {
				rk[sa[i]] = ++ tot;
			}
		}
		if(tot == n){
			break;
		}
		m = tot;
	}
	int k = 0;
	for(int i = 1; i <= n; ++ i){
		if(rk[i] == 1){
			h[1] = 0;
		}
		if(k){
			-- k;
		}
		int j = sa[rk[i]-1];
		while(j+k <= n && i+k <= n && s[i+k] == s[j+k]){
			++ k;
		}
		h[rk[i]] = k;
	}
}
 
int st[N][20];
void table(){
	for(int i = 1; i <= n; ++ i){
		st[i][0] = h[i];
	}
	for(int i = 1; i < 20; ++ i){
		for(int j = 1; j + (1 << i) - 1 <= n; ++ j){
			st[j][i] = min(st[j][i-1], st[j+(1<<i-1)][i-1]);
		}
	}
}
int qry(int l, int r){
	if(l == r){
		return 1e9;
	}
	if(l > r){
		swap(l, r);
	}
	++ l;
	int k = 31 ^ __builtin_clz(r - l + 1);
	return min(st[l][k], st[r-(1<<k)+1][k]);
}
 
int t[N*M], ls[N*M], rs[N*M], rt[N], tot;
int add(int p, int l, int r, int x){
	++ tot;
	tie(t[tot], ls[tot], rs[tot]) = tie(t[p], ls[p], rs[p]);
	p = tot;
	if(l == r){
		++ t[p];
	} else {
		int mid = l + r >> 1;
		if(x <= mid){
			ls[p] = add(ls[p], l, mid, x);
		} else {
			rs[p] = add(rs[p], mid+1, r, x);
		}
		t[p] = t[ls[p]] + t[rs[p]];
	}
	return p;
}
int sum(int p, int q, int l, int r, int ql, int qr){
	if(qr < l || r < ql){
		return 0;
	} else if(ql <= l && r <= qr){
		return t[p] - t[q];
	} else {
		int mid = l + r >> 1;
		return sum(ls[p], ls[q], l, mid, ql, qr) + sum(rs[p], rs[q], mid+1, r, ql, qr);
	}
}
void seg(){
	for(int i = 1; i <= n; ++ i){
		rt[i] = add(rt[i-1], 1, n, rk[i]);
	}
}
 
bool chk(int a, int b, int c, int d, int k){
	int l = a, r = b - k + 1;
	int x, y;
	int L = 1, R = rk[c];
	while(L < R){
		int mid = L + R >> 1;
		if(qry(mid, rk[c]) >= k){
			R = mid;
		} else {
			L = mid + 1;
		}
	}
	x = L;
	L = rk[c], R = n;
	while(L < R){
		int mid = L + R + 1 >> 1;
		if(qry(rk[c], mid) >= k){
			L = mid;
		} else {
			R = mid - 1;
		}
	}
	y = L;
	return sum(rt[r], rt[l-1], 1, n, x, y);
}
 
void solve(){
	scanf("%d%d", &n, &q);
	scanf("%s", s + 1);
	suf();
	seg();
	table();
	while(q--){
		int a, b, c, d;
		scanf("%d%d%d%d", &a, &b, &c, &d);
		int l = 0, r = min(d - c + 1, b - a + 1);
		while(l < r){
			int mid = l + r + 1 >> 1;
			if(chk(a, b, c, d, mid)){
				l = mid;
			} else {
				r = mid - 1;
			}
		}
		printf("%d\n", l);
	}
}

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posted @ 2024-04-03 20:10 KiharaTouma 阅读(41) 评论(1) 编辑收藏举报

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KiharaTouma

木原当麻

2024.04 别急记录

1. 餐巾计划问题

2. 国家集训队 - happiness

3. qoj1427 - Flip

4. USACO 2024 OPEN - Cowreography G

5. CCO2019 - Sirtet

6. LuoguP3600 - 随机数生成器

7. AGC066A - Adjacent Difference

8. AGC066B - Decreasing Digit Sums

9. AGC066C - Delete AAB or BAA

10. CF526F - Pudding Monsters

11. NOI2023 - 贸易

12. THUPC 2024 初赛 - 多折较差验证

13. AGC035C - Skolem XOR Tree

14. CF1874E - Jellyfish and Hack

15. AGC030C - Coloring Torus

16. NOI2020 - 超现实树

17. NOI2020 - 制作菜品

18. LuoguP2633 - Count on a tree

19. 国家集训队 - middle

20. HEOI/TJOI2016 - 字符串

最新随笔

推荐排行榜

	//P1251
	#include <bits/stdc++.h>
	using namespace std; typedef long long ll;
	void solve();int main(){ solve(); return 0; }

	const int N = 4e3 + 10, M = 4e4 + 10;
	const ll inf = 1e10;
	int n, r[N], p, u, f, v, s;
	ll mc, ln[M], cs[M], ds[N];
	int hd[N], eg[M], nx[M], tot = 1, nw[N], vs[N];

	void adg(int u, int v, ll w, ll c){
	eg[++tot] = v;
	ln[tot] = w;
	cs[tot] = c;
	nx[tot] = hd[u];
	hd[u] = tot;
	}
	void add(int u, int v, ll w, ll c){
	// printf("%d %d %lld %lld\n", u, v, w, c);
	adg(u, v, w, c);
	adg(v, u, 0, -c);
	}
	bool spfa(int s, int t){
	queue<int> q;
	memset(ds, 0x3f, sizeof(ds));
	memcpy(nw, hd, sizeof(hd));
	q.push(s);
	ds[s] = 0;
	vs[s] = 1;
	while(!q.empty()){
	int x = q.front();
	q.pop();
	vs[x] = 0;
	for(int i = hd[x]; i; i = nx[i]){
	int y = eg[i];
	ll z = ln[i], c = cs[i];
	if(z && ds[y] > ds[x] + c){
	ds[y] = ds[x] + c;
	if(!vs[y]){
	vs[y] = 1;
	q.push(y);
	}
	}
	}
	}
	return ds[t] != 0x3f3f3f3f3f3f3f3f;
	}
	ll dfs(int x, int t, ll fl){
	if(x == t){
	return fl;
	}
	ll rs = fl;
	vs[x] = 1;
	for(int i = nw[x]; i && rs; i = nx[i]){
	int y = eg[i];
	ll z = ln[i], c = cs[i];
	nw[x] = i;
	if(!vs[y] && z && ds[y] == ds[x] + c){
	ll k = dfs(y, t, min(rs, z));
	if(!k){
	ds[y] = 0;
	}
	ln[i] -= k;
	ln[i^1] += k;
	rs -= k;
	mc += k * c;
	}
	}
	vs[x] = 0;
	return fl - rs;
	}

	void solve(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++ i){
	scanf("%d", &r[i]);
	}
	scanf("%d%d%d%d%d", &p, &u, &f, &v, &s);
	int st = n + n + 1, ed = n + n + 2;
	for(int i = 1; i <= n; ++ i){
	add(i, i+n, r[i], 0);
	add(st, i+n, r[i], 0);
	add(i, ed, r[i], 0);
	if(i < n){
	add(i, i+1, inf, 0);
	}
	if(i + v <= n){
	add(i+n, i+v, inf, s);
	}
	if(i + u <= n){
	add(i+n, i+u, inf, f);
	}
	}
	add(st, 1, inf, p);
	while(spfa(st, ed)){
	while(dfs(st, ed, inf*2));
	}
	printf("%lld\n", mc);
	}

	//P1646
	#include <bits/stdc++.h>
	using namespace std;

	const int N = 5e4 + 10, inf = 100000, M = 3e5 + 10;
	int n, m, hd[N], eg[M], ln[M], nx[M], tot = 1, now[N], dep[N];

	void adg(int u, int v, int w){
	eg[++tot] = v;
	ln[tot] = w;
	nx[tot] = hd[u];
	hd[u] = tot;
	}
	void add(int u, int v, int w){
	adg(u, v, w);
	adg(v, u, 0);
	}
	bool bfs(int s, int t){
	queue<int> q;
	q.push(s);
	memcpy(now, hd, sizeof(hd));
	memset(dep, 0, sizeof(dep));
	dep[s] = 1;
	while(!q.empty()){
	int x = q.front();
	q.pop();
	for(int i = hd[x]; i; i = nx[i]){
	int y = eg[i], z = ln[i];
	if(z && !dep[y]){
	dep[y] = dep[x] + 1;
	q.push(y);
	if(y == t){
	return 1;
	}
	}
	}
	}
	return 0;
	}
	int dfs(int x, int t, int fl){
	if(x == t){
	return fl;
	} else {
	int rs = fl;
	for(int i = now[x]; i && rs; i = nx[i]){
	int y = eg[i], z = ln[i];
	now[x] = i;
	if(z && dep[y] == dep[x] + 1){
	int k = dfs(y, t, min(rs, z));
	if(!k){
	dep[y] = 0;
	}
	ln[i] -= k;
	ln[i^1] += k;
	rs -= k;
	}
	}
	return fl - rs;
	}
	}

	int main(){
	#define cg(x, y) ((x - 1) * m + y)
	scanf("%d%d", &n, &m);
	int sm = 0;
	int st = cg(n, m) + 1, ed = cg(n, m) + 2, a, cnt = cg(n, m) + 2;
	for(int i = 1; i <= n; ++ i){
	for(int j = 1; j <= m; ++ j){
	scanf("%d", &a);
	sm += a;
	add(st, cg(i, j), a);
	}
	}
	for(int i = 1; i <= n; ++ i){
	for(int j = 1; j <= m; ++ j){
	scanf("%d", &a);
	sm += a;
	add(cg(i, j), ed, a);
	}
	}
	for(int i = 1; i < n; ++ i){
	for(int j = 1; j <= m; ++ j){
	scanf("%d", &a);
	sm += a;
	++ cnt;
	add(st, cnt, a);
	add(cnt, cg(i, j), inf);
	add(cnt, cg(i+1, j), inf);
	}
	}
	for(int i = 1; i < n; ++ i){
	for(int j = 1; j <= m; ++ j){
	scanf("%d", &a);
	sm += a;
	++ cnt;
	add(cnt, ed, a);
	add(cg(i, j), cnt, inf);
	add(cg(i+1, j), cnt, inf);
	}
	}
	for(int i = 1; i <= n; ++ i){
	for(int j = 1; j < m; ++ j){
	scanf("%d", &a);
	sm += a;
	++ cnt;
	add(st, cnt, a);
	add(cnt, cg(i, j), inf);
	add(cnt, cg(i, j+1), inf);
	}
	}
	for(int i = 1; i <= n; ++ i){
	for(int j = 1; j < m; ++ j){
	scanf("%d", &a);
	sm += a;
	++ cnt;
	add(cnt, ed, a);
	add(cg(i, j), cnt, inf);
	add(cg(i, j+1), cnt, inf);
	}
	}
	int mf = 0, tmp;
	while(bfs(st, ed)){
	while(tmp = dfs(st, ed, inf)) mf += tmp;
	}
	printf("%d\n", sm - mf);
	return 0;
	}

	//qoj1427
	#include <bits/stdc++.h>
	using namespace std;
	typedef long long ll;

	const ll P = 998244353;
	const int N = 2e5 + 10;
	int n, m, c, a[N];
	ll fac[N], inv[N], pw[N], ivy, ans[N];
	vector<pair<int, int> > g[N], h[N];

	ll qp(ll x, ll y){
	ll ans = 1;
	while(y){
	if(y & 1){
	ans = ans * x % P;
	}
	x = x * x % P;
	y >>= 1;
	}
	return ans;
	}
	ll C(int x, int y){
	if(x < y) return 0;
	return fac[x] * inv[y] % P * inv[x-y] % P;
	}

	int main(){
	fac[0] = pw[0] = 1;
	for(int i = 1; i < N; ++ i){
	fac[i] = fac[i-1] * i % P;
	pw[i] = pw[i-1] * 2 % P;
	}
	inv[N-1] = qp(fac[N-1], P-2);
	for(int i = N - 2; i >= 0; -- i){
	inv[i] = inv[i+1] * (i+1) % P;
	}
	scanf("%d%d", &n, &m);
	ivy = qp(pw[n+n], P-2);
	for(int cs = 1; cs <= m; ++ cs){
	scanf("%d", &c);
	for(int i = 1; i <= c; ++ i){
	scanf("%d", &a[i]);
	}
	if(a[c] < n + n){
	ans[cs] = (ans[cs] + C(a[c]-c, n-c) * pw[n+n-a[c]]) % P;
	}
	g[c].push_back(make_pair(max(n, a[c]+1), cs));
	int ls = n - 1;
	if(c > 450){
	for(int i = 1; i <= c; ++ i){
	if(a[i] > ls){
	for(int k = ls+1; k < a[i]; ++ k){
	ans[cs] = (ans[cs] + C(k-i, n-1) * pw[n+n-k]) % P;
	}
	ls = a[i];
	}
	}
	h[c+1].push_back(make_pair(ls+1, cs));
	h[c+1].push_back(make_pair(n+n, -cs));
	} else {
	for(int i = 1; i <= c; ++ i){
	if(a[i] > ls){
	h[i].push_back(make_pair(ls+1, cs));
	h[i].push_back(make_pair(a[i], -cs));
	ls = a[i];
	}
	}
	h[c+1].push_back(make_pair(ls+1, cs));
	h[c+1].push_back(make_pair(n+n, -cs));
	}
	}
	for(int i = 1; i <= n; ++ i){
	if(g[i].size()){
	sort(g[i].begin(), g[i].end());
	reverse(g[i].begin(), g[i].end());
	int nw = n + n;
	ll rs = 0;
	for(auto j : g[i]){
	while(nw > j.first){
	-- nw;
	rs = (rs + C(nw-1-i, n-1-i) * pw[n+n-nw]) % P;
	}
	ans[j.second] = (ans[j.second] + rs) % P;
	}
	}
	}
	for(int i = 1; i <= n + 1; ++ i){
	if(h[i].size()){
	sort(h[i].begin(), h[i].end());
	reverse(h[i].begin(), h[i].end());
	int nw = n + n;
	ll rs = 0;
	for(auto j : h[i]){
	while(nw > j.first){
	-- nw;
	rs = (rs + C(nw-i, n-1) * pw[n+n-nw]) % P;
	}
	if(j.second > 0){
	ans[j.second] = (ans[j.second] + rs) % P;
	} else {
	ans[-j.second] = (ans[-j.second] - rs + P) % P;
	}
	}
	}
	}
	for(int i = 1; i <= m; ++ i){
	printf("%lld\n", 2 * ans[i] * ivy % P);
	}
	return 0;
	}

	//P10280
	#include <bits/stdc++.h>
	using namespace std;

	const int N = 1e6 + 10;
	int n, k;
	char s[N], t[N];
	long long ans;
	set<pair<int, int> > st[2];

	int main(){
	scanf("%d%d", &n, &k);
	scanf("%s%s", s+1, t+1);
	for(int i = 1; i <= n; ++ i){
	if(s[i] != t[i]){
	int x = s[i] - '0';
	if(st[!x].empty()){
	st[x].insert(make_pair(i%k, i));
	} else {
	auto it = st[!x].lower_bound(make_pair(i % k, 0));
	if(it == st[!x].end()){
	it = st[!x].begin();
	}
	ans += (i - (*it).second - 1) / k + 1;
	st[!x].erase(it);
	}
	}
	}
	printf("%lld\n", ans);
	return 0;
	}

	//P5532
	#include <bits/stdc++.h>
	using namespace std;
	typedef long long ll;

	const int N = 5e6 + 10;
	const int dx[4] = {0, -1, 1, 0}, dy[4] = {-1, 0, 0, 1};
	int plv[N], pcl[N], vs[N], ds[N], n, m, cs;
	vector<pair<int, int> > g[N];
	char pl[N], ans[N];

	void dfs(int x, int y, int l, int c){
	char (&s)[n+3][m+3] = decltype(s)(pl);
	int (&cl)[n+3][m+3] = decltype(cl)(pcl);
	int (&lv)[n+3][m+3] = decltype(lv)(plv);
	lv[x][y] = l;
	cl[x][y] = c;
	for(int i = 0; i < 4; ++ i){
	int xx = x + dx[i], yy = y + dy[i];
	if(s[xx][yy] == '#' && !lv[xx][yy]){
	dfs(xx, yy, l-dx[i], c);
	}
	}
	}
	void spfa(int st){
	queue<int> q;
	memset(vs, 0, sizeof(vs));
	memset(ds, 0x3f, sizeof(ds));
	vs[st] = 1, ds[st] = 0, q.push(st);
	while(!q.empty()){
	int x = q.front();
	q.pop();
	vs[x] = 0;
	for(auto i : g[x]){
	if(ds[i.first] > ds[x] + i.second){
	ds[i.first] = ds[x] + i.second;
	if(!vs[i.first]){
	q.push(i.first);
	vs[i.first] = 1;
	}
	}
	}
	}
	}

	int main(){
	scanf("%d%d", &n, &m);
	char (&s)[n+3][m+3] = decltype(s)(pl);
	char (&t)[n+3][m+3] = decltype(t)(ans);
	int (&cl)[n+3][m+3] = decltype(cl)(pcl);
	int (&lv)[n+3][m+3] = decltype(lv)(plv);
	for(int i = 1; i <= n; ++ i){
	scanf("%s", s[i] + 1);
	}
	for(int i = n; i >= 1; -- i){
	for(int j = 1; j <= m; ++ j){
	t[i][j] = '.';
	if(!lv[i][j] && s[i][j] == '#'){
	dfs(i, j, 1, ++ cs);
	}
	}
	}
	for(int j = 1; j <= m; ++ j){
	int ls = 0;
	for(int i = n; i >= 1; -- i){
	if(s[i][j] != '#'){
	continue;
	}
	if(ls && cl[i][j] != cl[ls][j]){
	g[cl[ls][j]].emplace_back(cl[i][j], lv[i][j]-lv[ls][j]-1);
	}
	ls = i;
	}
	}
	for(int i = 1; i <= cs; ++ i){
	g[cs+1].emplace_back(i, 0);
	}
	spfa(cs + 1);
	for(int i = 1; i <= n; ++ i){
	for(int j = 1; j <= m; ++ j){
	if(s[i][j] == '#'){
	t[ds[cl[i][j]]+n-lv[i][j]+1][j] = '#';
	}
	}
	}
	for(int i = 1; i <= n; ++ i){
	printf("%s\n", t[i] + 1);
	}
	return 0;
	}

	//P3600
	#include <bits/stdc++.h>
	using namespace std;
	typedef long long ll;

	const int N = 2010;
	const ll P = 666623333;
	int n, x, q, mr[N];
	ll f[N][N], g[N];

	ll qp(ll x, ll y){
	ll ans = 1;
	while(y){
	if(y & 1){
	ans = ans * x % P;
	}
	x = x * x % P;
	y >>= 1;
	}
	return ans;
	}

	ll t[N4], tag[N4];
	void psd(int p){
	t[p<<1] = t[p<<1] * tag[p] % P;
	t[p<<1\|1] = t[p<<1\|1] * tag[p] % P;
	tag[p<<1] = tag[p<<1] * tag[p] % P;
	tag[p<<1\|1] = tag[p<<1\|1] * tag[p] % P;
	tag[p] = 1;
	}
	void mul(int p, int l, int r, int ql, int qr, ll v){
	if(ql <= l && r <= qr){
	t[p] = t[p] * v % P;
	tag[p] = tag[p] * v % P;
	} else if(r < ql \|\| qr < l){
	return;
	} else {
	int mid = l + r >> 1;
	psd(p);
	mul(p<<1, l, mid, ql, qr, v);
	mul(p<<1\|1, mid+1, r, ql, qr, v);
	t[p] = (t[p<<1] + t[p<<1\|1]) % P;
	}
	}
	void add(int p, int l, int r, int x, ll v){
	if(l == r){
	t[p] = (t[p] + v) % P;
	} else {
	int mid = l + r >> 1;
	psd(p);
	if(x <= mid){
	add(p<<1, l, mid, x, v);
	} else {
	add(p<<1\|1, mid+1, r, x, v);
	}
	t[p] = (t[p<<1] + t[p<<1\|1]) % P;
	}
	}
	ll qry(int p, int l, int r, int ql, int qr){
	if(ql <= l && r <= qr){
	return t[p];
	} else if(r < ql \|\| qr < l){
	return 0;
	} else {
	int mid = l + r >> 1;
	psd(p);
	return (qry(p<<1, l, mid, ql, qr) +
	qry(p<<1\|1, mid+1, r, ql, qr)) % P;
	}
	}

	int main(){
	scanf("%d%d%d", &n, &x, &q);
	for(int i = 1; i <= q; ++ i){
	int l, r;
	scanf("%d%d", &l, &r);
	mr[r] = max(mr[r], l);
	}
	for(int T = 1; T <= x; ++ T){
	mul(1, 0, n, 0, n, 0);
	add(1, 0, n, 0, 1);
	for(int i = 1; i <= n; ++ i){
	ll val = t[1] * T % P;
	t[1] = t[1] * (x-T) % P;
	tag[1] = tag[1] * (x-T) % P;
	mul(1, 0, n, 0, mr[i]-1, 0);
	add(1, 0, n, i, val);
	}
	g[T] = t[1];
	}
	ll ans = 0;
	for(int i = n; i >= 1; -- i){
	g[i] = (g[i] - g[i-1] + P) % P;
	ans += g[i] * i % P;
	ans %= P;
	}
	printf("%lld\n", ans * qp(qp(x, n), P-2) % P);
	return 0;
	}

	//AT_agc066_a
	#include <bits/stdc++.h>
	using namespace std; typedef long long ll;
	void solve();int main(){ solve(); return 0; }

	const int N = 510;
	int n, d, a[N][N], b[N][N], c[N][N], sb, sc;

	void solve(){
	scanf("%d%d", &n, &d);
	for(int i = 1; i <= n; ++ i){
	for(int j = 1; j <= n; ++ j){
	scanf("%d", &a[i][j]);
	a[i][j] += 1000;
	if(((i + j) & 1) ^ ((a[i][j] / d) & 1)){
	b[i][j] = a[i][j] - a[i][j] % d;
	c[i][j] = b[i][j] + d;
	sb += a[i][j] - b[i][j];
	sc += c[i][j] - a[i][j];
	} else {
	c[i][j] = a[i][j] - a[i][j] % d;
	b[i][j] = c[i][j] + d;
	sc += a[i][j] - c[i][j];
	sb += b[i][j] - a[i][j];
	}
	}
	}
	for(int i = 1; i <= n; ++ i){
	for(int j = 1; j <= n; ++ j){
	printf("%d ", (sb < sc ? b[i][j] : c[i][j]) - 1000);
	}
	puts("");
	}
	}

	//AT_agc066_b
	#include <bits/stdc++.h>
	using namespace std; typedef long long ll;
	void solve();int main(){ solve(); return 0; }

	const int N = 10010;
	int s[N], t[N];
	int n, len = 1, tot = 1;

	void solve(){
	cin >> n;
	t[1] = s[1] = 5;
	for(int i = 2; i <= 160; ++ i){
	for(int j = 1; j <= len; ++ j){
	t[j] = t[j] * 5;
	t[j] = t[j] + t[j-1] / 10;
	t[j-1] %= 10;
	}
	while(t[len] >= 10){
	t[len+1] = t[len] / 10;
	t[len] %= 10;
	++ len;
	}
	for(int j = 1; j <= len; ++ j){
	s[++tot] = t[j];
	}
	}
	for(int i = tot; i >= 1; -- i){
	printf("%d", s[i]);
	}
	puts("");
	}

	//AT_agc066_c
	#include <bits/stdc++.h>
	using namespace std; typedef long long ll;
	void solve();int main(){ solve(); return 0; }

	const int N = 5e6 + 10;
	int T, n, f[N], g[N], h[N];
	char s[N];

	void solve(){
	scanf("%d", &T);
	while(T--){
	scanf("%s", s + 1);
	n = strlen(s + 1);
	int sum = N / 2;
	for(int i = sum - n2; i <= sum + n2; ++ i){
	g[i] = h[i] = 0x3f3f3f3f;
	}
	for(int i = 0; i <= n; ++ i){
	if(i){
	f[i] = f[i-1] + 1;
	if(s[i] == 'A'){
	++ sum;
	f[i] = min(f[i], g[sum]);
	} else {
	-- sum;
	-- sum;
	f[i] = min(f[i], g[sum]);
	f[i] = min(f[i], h[sum]);
	}
	}
	if(s[i+1] == 'B'){
	g[sum] = min(g[sum], f[i]);
	} else {
	h[sum] = min(h[sum], f[i]);
	}
	}
	printf("%d\n", (n - f[n]) / 3);
	for(int i = 0; i <= n; ++ i){
	f[i] = 0;
	}
	}
	}

	//CF526F
	#include <bits/stdc++.h>
	using namespace std; typedef long long ll;
	void solve();int main(){ solve(); return 0; }

	const int N = 3e5 + 10;
	int n, a[N], mx[N], mn[N], tx, tn;
	ll ans = 0;

	struct node{
	int tag, mn, cnt;
	} t[N*4];

	void psd(int p){
	t[p<<1].tag += t[p].tag;
	t[p<<1].mn += t[p].tag;
	t[p<<1\|1].tag += t[p].tag;
	t[p<<1\|1].mn += t[p].tag;
	t[p].tag = 0;
	}

	void build(int p, int l, int r){
	t[p].cnt = r - l + 1;
	if(l == r){
	return;
	}
	int mid = l + r >> 1;
	build(p<<1, l, mid);
	build(p<<1\|1, mid+1, r);
	}
	void add(int p, int l, int r, int ql, int qr, int x){
	if(qr < l \|\| r < ql){
	return;
	} else if(ql <= l && r <= qr){
	t[p].mn += x;
	t[p].tag += x;
	} else {
	int mid = l + r >> 1;
	psd(p);
	add(p<<1, l, mid, ql, qr, x);
	add(p<<1\|1, mid+1, r, ql, qr, x);
	t[p].mn = min(t[p<<1].mn, t[p<<1\|1].mn);
	if(t[p<<1].mn < t[p<<1\|1].mn){
	t[p].cnt = t[p<<1].cnt;
	} else if(t[p<<1].mn > t[p<<1\|1].mn){
	t[p].cnt = t[p<<1\|1].cnt;
	} else {
	t[p].cnt = t[p<<1].cnt + t[p<<1\|1].cnt;
	}
	}
	}
	pair<int, int> qry(int p, int l, int r, int ql, int qr){
	if(qr < l \|\| r < ql){
	return make_pair(1e9, 0);
	} else if(ql <= l && r <= qr){
	return make_pair(t[p].mn, t[p].cnt);
	} else {
	int mid = l + r >> 1;
	psd(p);
	pair<int, int> x = qry(p<<1, l, mid, ql, qr);
	pair<int, int> y = qry(p<<1\|1, mid+1, r, ql, qr);
	if(x.first < y.first){
	return x;
	} else if(x.first > y.first){
	return y;
	} else {
	return make_pair(x.first, x.second + y.second);
	}
	}
	}

	void solve(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++ i){
	int x, y;
	scanf("%d%d", &x, &y);
	a[x] = y;
	}
	build(1, 1, n);
	for(int i = 1; i <= n; ++ i){
	add(1, 1, n, i, i, i);
	while(tx && a[mx[tx]] < a[i]){
	add(1, 1, n, mx[tx-1]+1, mx[tx], -a[mx[tx]]);
	-- tx;
	}
	add(1, 1, n, mx[tx]+1, i, a[i]);
	mx[++tx] = i;
	while(tn && a[mn[tn]] > a[i]){
	add(1, 1, n, mn[tn-1]+1, mn[tn], a[mn[tn]]);
	-- tn;
	}
	add(1, 1, n, mn[tn]+1, i, -a[i]);
	mn[++tn] = i;
	ans += qry(1, 1, n, 1, i).second;
	}
	printf("%lld\n", ans);
	}

	//P9481
	//You can get an NOI2023 Ag by solving this problem!
	//You can get an NOI2023 Au by 509pts.
	//You can get 5pts by O(3^n) 后面忘了
	//509+5=514
	//then 进入集训队！！！
	#include <bits/stdc++.h>
	using namespace std;
	typedef long long ll;

	const int N = 262150;
	const ll P = 998244353;
	int n, m, a[N], dep[N], vis[N];
	vector<pair<int, int> > g[N];
	vector<int> eg[N], son[N];
	ll sum[N], dis[N], ans, fa[N][20];
	struct edge{
	int u, v, w;
	} e[N];

	int main(){
	scanf("%d%d", &n, &m);
	n = (1 << n) - 1;
	dep[1] = 1;
	son[1].push_back(1);
	for(int i = 2; i <= n; ++ i){
	son[i].push_back(i);
	scanf("%d", &a[i]);
	dep[i] = dep[i>>1] + 1;
	for(int j = 1; j <= dep[i]; ++ j){
	fa[i][j] = fa[i][j-1] + a[i>>(j-1)];
	sum[i>>j] = (sum[i>>j] + fa[i][j]) % P;
	son[i>>j].push_back(i);
	}
	}
	for(int i = 1; i <= m; ++ i){
	int u, v, w;
	scanf("%d%d%d", &u, &v, &w);
	e[i] = { u, v, w };
	g[u].emplace_back(v, w);
	int x = v;
	while(x != u){
	eg[x].push_back(i);
	x >>= 1;
	}
	}
	for(int rt = 1; rt <= n / 2; ++ rt){
	priority_queue<pair<ll, int> > q;
	for(int j : son[rt]){
	dis[j] = 1e18;
	vis[j] = 0;
	}
	dis[rt] = 0;
	q.push(make_pair(-dis[rt], rt));
	while(!q.empty()){
	int x = q.top().second;
	q.pop();
	if(vis[x] == 1){
	continue;
	}
	vis[x] = 1;
	if(x == rt){
	for(int i : eg[x]){
	int y = e[i].v;
	ll z = e[i].w;
	z += fa[x][dep[x] - dep[e[i].u]];
	if(dis[y] > dis[x] + z){
	dis[y] = dis[x] + z;
	q.push(make_pair(-dis[y], y));
	}
	}
	}
	for(auto i : g[x]){
	int y = i.first;
	ll z = i.second;
	if(dis[y] > dis[x] + z){
	dis[y] = dis[x] + z;
	q.push(make_pair(-dis[y], y));
	}
	}
	if((x>>1) >= rt && dis[x>>1] > dis[x] + a[x]){
	dis[x>>1] = dis[x] + a[x];
	q.push(make_pair(-dis[x>>1], x>>1));
	}
	}
	ll le = 0, cle = 1;
	for(int j : son[rt<<1]){
	if(dis[j] > 1e16){
	continue;
	}
	++ cle;
	le = (le + dis[j]) % P;
	}
	ans = (ans + cle * (sum[rt<<1\|1] + a[rt<<1\|1] * son[rt<<1\|1].size() % P)) % P;
	ans = (ans + le * (son[rt<<1\|1].size() + 1)) % P;
	cle = 1, le = 0;
	for(int j : son[rt<<1\|1]){
	if(dis[j] > 1e16){
	continue;
	}
	++ cle;
	le = (le + dis[j]) % P;
	}
	ans = (ans + cle * (sum[rt<<1] + a[rt<<1] * son[rt<<1].size() % P)) % P;
	ans = (ans + le * (son[rt<<1].size() + 1)) % P;
	}
	printf("%lld\n", ans);
	return 0;
	}

	//P9964
	#include <bits/stdc++.h>
	using namespace std;

	const int N = 5010;
	int n, f[N][N][2], ok[N][N], g[N][N][2];
	char s[N];

	int main(){
	scanf("%d%s", &n, s+1);
	for(int len = 1; len <= n; len += 2){
	for(int l = 1; l + len - 1 <= n; ++ l){
	int r = l + len - 1;
	if(l == r){
	ok[l][r] = 1;
	} else {
	ok[l][r] = (s[l] != s[r]) & ok[l+1][r-1];
	}
	if(ok[l][r]){
	g[l][r][0] = g[l][r][1] = (l+r) >> 1;
	}
	}
	}
	memset(f, 0x3f, sizeof(f));
	for(int i = 1; i <= n; ++ i){
	for(int j = i; j <= n; ++ j){
	if(g[i][j][0] == 0){
	g[i][j][0] = g[i][j-1][0];
	}
	}
	}
	for(int i = n; i >= 1; -- i){
	for(int j = i; j >= 1; -- j){
	if(g[j][i][1] == 0){
	g[j][i][1] = g[j+1][i][1];
	}
	}
	}
	for(int len = 1; len <= n; ++ len){
	for(int l = 1; l + len - 1 <= n; ++ l){
	int r = l + len - 1;
	if(l == r){
	f[l][r][0] = 1;
	f[l][r][1] = 0;
	} else {
	int x = g[l][r][0], y = g[l][r][1];
	if(f[x+1][r][0] + 1 < f[l][r][0]){
	f[l][r][0] = f[x+1][r][0] + 1;
	f[l][r][1] = f[x+1][r][1] + r + l - x - x;
	} else if(f[x+1][r][0] + 1 == f[l][r][0]){
	f[l][r][1] = min(f[l][r][1], f[x+1][r][1] + r + l - x - x);
	}
	if(f[l][y-1][0] + 1 < f[l][r][0]){
	f[l][r][0] = f[l][y-1][0] + 1;
	f[l][r][1] = f[l][y-1][1] + y + y - r - l;
	} else if(f[l][y-1][0] + 1 == f[l][r][0]){
	f[l][r][1] = min(f[l][r][1], f[l][y-1][1] + y + y - r - l);
	}
	}
	}
	}
	printf("%d %d\n", f[1][n][0], f[1][n][1]);
	return 0;
	}

	//AT_agc035_c
	#include <bits/stdc++.h>
	using namespace std;

	const int N = 1e5 + 10;
	int n;

	int main(){
	scanf("%d", &n);
	int p = 1;
	while(p < n){
	p *= 2;
	}
	if(p == n){
	puts("No");
	} else {
	puts("Yes");
	printf("%d %d\n", 1+n, 3);
	for(int i = 2; i <= n - 1; i += 2){
	printf("%d %d\n", i, 1);
	printf("%d %d\n", i, i+1);
	printf("%d %d\n", i+n+1, 1);
	printf("%d %d\n", i+n+1, i+n);
	}
	if(n % 2 == 0){
	printf("%d %d\n", n, n-2);
	printf("%d %d\n", n+n, n+(n^(n-2)^1));
	}
	}
	return 0;
	}

	//CF1874E
	#include <bits/stdc++.h>
	using namespace std; typedef long long ll;
	void solve();int main(){ solve(); return 0; }

	const int N = 210;
	const ll P = 1e9 + 7;
	ll y[N*N], C[N][N], F[N];
	ll fac[NN], f[NN], p[NN], q[NN];
	int n, lim, m;

	ll qp(ll x, ll y){
	ll ans = 1;
	while(y){
	if(y & 1){
	ans = ans * x % P;
	}
	x = x * x % P;
	y >>= 1;
	}
	return ans;
	}

	void solve(){
	cin >> n >> lim;
	m = n * (n + 1) / 2 + 1;
	for(int i = 0; i <= n; ++ i){
	C[i][0] = C[i][i] = 1;
	for(int j = 1; j < i; ++ j){
	C[i][j] = (C[i-1][j] + C[i-1][j-1]) % P;
	}
	}
	fac[0] = 1;
	for(int i = 1; i <= m; ++ i){
	fac[i] = fac[i-1] * i % P;
	F[0] = 1;
	ll pw = 1;
	for(int k = 1; k <= n; ++ k){
	pw = pw * i % P;
	F[k] = 0;
	for(int j = 1; j <= k; ++ j){
	F[k] = (F[k] + C[k-1][j-1] * F[j-1] % P * F[k-j]) % P;
	}
	F[k] = F[k] * pw % P;
	}
	y[i] = F[n];
	}
	p[0] = 1;
	for(int i = 1; i <= m; ++ i){
	for(int j = 1; j <= i; ++ j){
	q[j] = p[j-1];
	p[j-1] = p[j-1] * (P - i) % P;
	}
	for(int j = 0; j <= i; ++ j){
	p[j] = (p[j] + q[j]) % P;
	}
	}
	for(int i = 1; i <= m; ++ i){
	ll val = y[i] * qp(fac[i-1], P-2) % P * qp(fac[m-i], P-2) % P;
	if((m - i) & 1){
	val = P - val;
	}
	for(int j = 0; j <= m; ++ j){
	q[j] = p[j];
	}
	for(int j = m - 1; j >= 0; -- j){
	ll r = q[j+1];
	f[j] = (f[j] + r * val) % P;
	q[j] = (q[j] + r * i) % P;
	}
	}
	ll ans = 0;
	for(int i = lim; i < m; ++ i){
	ans = (ans + f[i]) % P;
	}
	cout << ans << '\n';
	}

	//AT_agc030_c
	#include <bits/stdc++.h>
	using namespace std; typedef long long ll;
	void solve();int main(){ solve(); return 0; }

	const int N = 510;
	int k, a[N][N];

	void solve(){
	// freopen(".out", "w", stdout);
	scanf("%d", &k);
	if(k <= 500){
	printf("%d\n", k);
	for(int i = 1; i <= k; ++ i){
	for(int j = 1; j <= k; ++ j){
	printf("%d ", j);
	}
	puts("");
	}
	} else {
	puts("500");
	for(int i = 1; i <= 500; ++ i){
	for(int j = 1, p = i; j <= 500; ++ j){
	a[j][p] = i;
	if(i + 500 <= k && j % 2 == 0){
	a[j][p] = i + 500;
	}
	++ p;
	if(p > 500){
	p -= 500;
	}
	}
	}
	for(int i = 1; i <= 500; ++ i){
	for(int j = 1; j <= 500; ++ j){
	printf("%d ", a[i][j]);
	}
	puts("");
	}
	}
	}

	//P6776
	#include <bits/stdc++.h>
	using namespace std;

	const int N = 2e6 + 10;
	int T, m, n;
	vector<pair<int, int> > g[N];
	int ch[N][4], tot = 1, ok[N];

	int sc(pair<int, int> x){
	return (x.first ? 1 : 0) + (x.second ? 1 : 0);
	}
	void dfs(int i, int p, int q){
	if(g[i][q].first && g[i][q].second){
	if(!sc(g[i][g[i][q].first])){
	if(!ch[p][2]) ch[p][2] = ++ tot;
	dfs(i, ch[p][2], g[i][q].second);
	}
	if(!sc(g[i][g[i][q].second])){
	if(!ch[p][3]) ch[p][3] = ++ tot;
	dfs(i, ch[p][3], g[i][q].first);
	}
	} else if(g[i][q].first){
	if(!ch[p][0]) ch[p][0] = ++ tot;
	dfs(i, ch[p][0], g[i][q].first);
	} else if(g[i][q].second){
	if(!ch[p][1]) ch[p][1] = ++ tot;
	dfs(i, ch[p][1], g[i][q].second);
	} else {
	ok[p] = 1;
	}
	}

	int main(){
	scanf("%d", &T);
	while(T--){
	scanf("%d", &m);
	for(int i = 1; i <= m; ++ i){
	scanf("%d", &n);
	g[i].resize(n + 1);
	for(int j = 1; j <= n; ++ j){
	scanf("%d%d", &g[i][j].first, &g[i][j].second);
	}
	bool flg = 1;
	for(int j = 1; j <= n; ++ j){
	if(sc(g[i][g[i][j].first]) && sc(g[i][g[i][j].second])){
	flg = 0;
	break;
	}
	}
	if(flg){
	dfs(i, 1, 1);
	}
	}
	for(int i = tot; i >= 1; -- i){
	if(ok[ch[i][0]] && ok[ch[i][1]] && ok[ch[i][2]] && ok[ch[i][3]]){
	ok[i] = 1;
	}
	}
	puts(ok[1] ? "Almost Complete" : "No");
	for(int i = 0; i <= tot; ++ i){
	ch[i][0] = ch[i][1] = ch[i][2] = ch[i][3] = ok[i] = 0;
	}
	tot = 1;
	}
	return 0;
	}

	//P6775
	#include <bits/stdc++.h>
	using namespace std;

	const int N = 510;
	int T, n, m, k, ok[N];
	pair<int, int> d[N], p[N];
	bitset<NN20> b[N];

	void solve(int cnt, int l, int r){
	int tp = l;
	for(int i = 1; i <= cnt; ++ i){
	sort(d + tp, d + r + 1);
	if(d[tp].first > k){
	printf("%d %d", d[tp].second, k);
	d[tp].first -= k;
	} else if(d[tp].first == k){
	printf("%d %d", d[tp].second, k);
	d[tp].first = 0;
	++ tp;
	} else {
	printf("%d %d ", d[tp].second, d[tp].first);
	printf("%d %d", d[r].second, k - d[tp].first);
	d[r].first -= (k - d[tp].first);
	d[tp].first = 0;
	++ tp;
	}
	puts("");
	}
	}

	int main(){
	scanf("%d", &T);
	while(T--){
	scanf("%d%d%d", &n, &m, &k);
	for(int i = 1; i <= n; ++ i){
	scanf("%d", &d[i].first);
	d[i].second = i;
	}
	if(m >= n - 1){
	solve(m, 1, n);
	} else {
	int tmp = n * k + 3;
	b[0].reset();
	b[0].set(tmp);
	for(int i = 1; i <= n; ++ i){
	if(d[i].first >= k){
	b[i] = b[i-1] \| (b[i-1] << d[i].first - k);
	} else {
	b[i] = b[i-1] \| (b[i-1] >> k - d[i].first);
	}
	}
	if(b[n].test(tmp - k)){
	int flg = tmp - k;
	int l = 0, r = 0;
	for(int i = n; i >= 1; -- i){
	if(b[i-1].test(flg)){
	continue;
	}
	for(int j = 1; j <= n; ++ j){
	if(!ok[i] && b[i-1].test(flg-d[i].first+k)){
	ok[i] = 1;
	flg -= d[i].first-k;
	++ r;
	}
	}
	}
	for(int i = 1; i <= n; ++ i){
	if(ok[i] == 1){
	p[++l] = d[i];
	} else {
	p[++r] = d[i];
	}
	ok[i] = 0;
	}
	memcpy(d, p, sizeof(p));
	solve(l-1, 1, l);
	solve(r-l-1, l+1, r);
	} else {
	puts("-1");
	}
	}
	}
	return 0;
	}

	//P2633
	#include <bits/stdc++.h>
	using namespace std; typedef long long ll;
	void solve();int main(){ solve(); return 0; }

	const int N = 1e5 + 10;
	int n, m, rg, a[N], b[N], anc[N][20], dep[N];
	int t[N20], ls[N20], rs[N*20], rt[N], cnt = 1;
	vector<int> g[N];

	int add(int p, int l, int r, int x){
	int nw = ++ cnt;
	tie(t[nw], ls[nw], rs[nw]) = tie(t[p], ls[p], rs[p]);
	if(l == r){
	++ t[nw];
	} else {
	int mid = l + r >> 1;
	if(x <= mid){
	ls[nw] = add(ls[nw], l, mid, x);
	} else {
	rs[nw] = add(rs[nw], mid+1, r, x);
	}
	t[nw] = t[ls[nw]] + t[rs[nw]];
	}
	return nw;
	}
	int qry(int p, int q, int x, int y, int l, int r, int k){
	if(l == r){
	return l;
	} else {
	int mid = l + r >> 1;
	int val = t[ls[p]] + t[ls[q]] - t[ls[x]] - t[ls[y]];
	if(val >= k){
	return qry(ls[p], ls[q], ls[x], ls[y], l, mid, k);
	} else {
	return qry(rs[p], rs[q], rs[x], rs[y], mid+1, r, k-val);
	}
	}
	}

	void dfs(int x, int fa){
	anc[x][0] = fa;
	dep[x] = dep[fa] + 1;
	rt[x] = add(rt[fa], 1, rg, a[x]);
	for(int i = 1; i < 20; ++ i){
	anc[x][i] = anc[anc[x][i-1]][i-1];
	}
	for(int i : g[x]){
	if(i != fa){
	dfs(i, x);
	}
	}
	}
	int lca(int x, int y){
	if(dep[x] < dep[y]){
	swap(x, y);
	}
	for(int i = 19; i >= 0; -- i){
	if(dep[anc[x][i]] >= dep[y]){
	x = anc[x][i];
	}
	}
	if(x == y){
	return x;
	}
	for(int i = 19; i >= 0; -- i){
	if(anc[x][i] != anc[y][i]){
	x = anc[x][i];
	y = anc[y][i];
	}
	}
	return anc[x][0];
	}

	void solve(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; ++ i){
	scanf("%d", &a[i]);
	b[i] = a[i];
	}
	sort(b + 1, b + n + 1);
	rg = unique(b + 1, b + n + 1) - b - 1;
	for(int i = 1; i <= n; ++ i){
	a[i] = lower_bound(b + 1, b + rg + 1, a[i]) - b;
	}
	for(int i = 1; i < n; ++ i){
	int x, y;
	scanf("%d%d", &x, &y);
	g[x].push_back(y);
	g[y].push_back(x);
	}
	rt[0] = ++ cnt;
	dfs(1, 0);
	int ls = 0;
	while(m--){
	int u, v, k;
	scanf("%d%d%d", &u, &v, &k);
	u ^= ls;
	int p = lca(u, v);
	ls = b[qry(rt[u], rt[v], rt[p], rt[anc[p][0]], 1, rg, k)];
	printf("%d\n", ls);
	}
	}

	//P2839
	#include <bits/stdc++.h>
	using namespace std; typedef long long ll;
	void solve();int main(){ solve(); return 0; }

	const int N = 20010, M = 400;
	int lrt[N], rrt[N], t[NM], tag[NM], ls[NM], rs[NM], cnt;
	int n, a[N], q, la, pos[N];
	pair<int, int> b[N];

	int build(int l, int r, int op){
	int p = ++ cnt;
	if(l == r){
	t[p] = op ? l : n - l + 1;
	} else {
	int mid = l + r >> 1;
	ls[p] = build(l, mid, op);
	rs[p] = build(mid+1, r, op);
	t[p] = max(t[ls[p]], t[rs[p]]);
	}
	return p;
	}
	int nnd(int p){
	++ cnt;
	tie(t[cnt], tag[cnt], ls[cnt], rs[cnt]) = tie(t[p], tag[p], ls[p], rs[p]);
	return cnt;
	}
	void psd(int p){
	if(!tag[p]){
	return;
	}
	ls[p] = nnd(ls[p]);
	rs[p] = nnd(rs[p]);
	tag[ls[p]] += tag[p];
	tag[rs[p]] += tag[p];
	t[ls[p]] += tag[p];
	t[rs[p]] += tag[p];
	tag[p] = 0;
	}
	int add(int p, int l, int r, int ql, int qr){
	p = nnd(p);
	if(ql <= l && r <= qr){
	tag[p] -= 2;
	t[p] -= 2;
	} else {
	int mid = l + r >> 1;
	psd(p);
	if(ql <= mid){
	ls[p] = add(ls[p], l, mid, ql, qr);
	}
	if(mid < qr){
	rs[p] = add(rs[p], mid+1, r, ql, qr);
	}
	t[p] = max(t[ls[p]], t[rs[p]]);
	}
	return p;
	}
	int ask(int p, int l, int r, int ql, int qr){
	if(ql <= l && r <= qr){
	return t[p];
	} else {
	int mid = l + r >> 1;
	psd(p);
	int ans = -2e9;
	if(ql <= mid){
	ans = max(ans, ask(ls[p], l, mid, ql, qr));
	}
	if(mid < qr){
	ans = max(ans, ask(rs[p], mid+1, r, ql, qr));
	}
	return ans;
	}
	}

	bool chk(int a, int b, int c, int d, int k){
	-- k;
	int val = 0;
	val += ask(lrt[k], 1, n, c, d) - ask(lrt[k], 1, n, b, b);
	val += ask(rrt[k], 1, n, a, b) - ask(rrt[k], 1, n, b+1, b+1);
	return val >= 0;
	}

	void solve(){
	scanf("%d", &n);
	for(int i = 1; i <= n; ++ i){
	scanf("%d", &a[i]);
	b[i] = make_pair(a[i], i);
	}
	sort(b + 1, b + n + 1);
	for(int i = 1; i <= n; ++ i){
	a[i] = lower_bound(b + 1, b + n + 1, make_pair(a[i], i)) - b;
	pos[a[i]] = i;
	}
	lrt[0] = build(1, n, 1);
	rrt[0] = build(1, n, 0);
	for(int i = 1; i <= n; ++ i){
	lrt[i] = add(lrt[i-1], 1, n, pos[i], n);
	rrt[i] = add(rrt[i-1], 1, n, 1, pos[i]);
	}
	scanf("%d", &q);
	while(q--){
	int p[4];
	scanf("%d%d%d%d", &p[0], &p[1], &p[2], &p[3]);
	p[0] = (p[0] + la) % n;
	p[1] = (p[1] + la) % n;
	p[2] = (p[2] + la) % n;
	p[3] = (p[3] + la) % n;
	sort(p, p + 4);
	int l = 1, r = n;
	while(l < r){
	int mid = l + r + 1 >> 1;
	if(chk(p[0] + 1, p[1] + 1, p[2] + 1, p[3] + 1, mid)){
	l = mid;
	} else {
	r = mid - 1;
	}
	}
	la = b[l].first;
	printf("%d\n", la);
	}
	}

	//P4094
	#include <bits/stdc++.h>
	using namespace std; typedef long long ll;
	void solve();int main(){ solve(); return 0; }

	const int N = 1e5 + 10, M = 30;
	int n, q;

	int sa[N], rk[N], cnt[N], pre[N], h[N];
	char s[N];
	void suf(){
	int m = 26;
	for(int i = 1; i <= n; ++ i){
	rk[i] = s[i] - 'a' + 1;
	++ cnt[rk[i]];
	}
	for(int i = 2; i <= m; ++ i){
	cnt[i] += cnt[i-1];
	}
	for(int i = n; i >= 1; -- i){
	sa[cnt[rk[i]]--] = i;
	}
	for(int k = 1; k <= n; k <<= 1){
	int tot = 0;
	for(int i = n - k + 1; i <= n; ++ i){
	pre[++tot] = i;
	}
	for(int i = 1; i <= n; ++ i){
	if(sa[i] > k){
	pre[++tot] = sa[i] - k;
	}
	}
	for(int i = 1; i <= m; ++ i){
	cnt[i] = 0;
	}
	for(int i = 1; i <= n; ++ i){
	++ cnt[rk[i]];
	}
	for(int i = 2; i <= m; ++ i){
	cnt[i] += cnt[i-1];
	}
	for(int i = n; i >= 1; -- i){
	sa[cnt[rk[pre[i]]]--] = pre[i];
	pre[i] = 0;
	}
	swap(rk, pre);
	tot = 1, rk[sa[1]] = 1;
	for(int i = 2; i <= n; ++ i){
	if(pre[sa[i]] == pre[sa[i-1]] && pre[sa[i]+k] == pre[sa[i-1]+k]){
	rk[sa[i]] = tot;
	} else {
	rk[sa[i]] = ++ tot;
	}
	}
	if(tot == n){
	break;
	}
	m = tot;
	}
	int k = 0;
	for(int i = 1; i <= n; ++ i){
	if(rk[i] == 1){
	h[1] = 0;
	}
	if(k){
	-- k;
	}
	int j = sa[rk[i]-1];
	while(j+k <= n && i+k <= n && s[i+k] == s[j+k]){
	++ k;
	}
	h[rk[i]] = k;
	}
	}

	int st[N][20];
	void table(){
	for(int i = 1; i <= n; ++ i){
	st[i][0] = h[i];
	}
	for(int i = 1; i < 20; ++ i){
	for(int j = 1; j + (1 << i) - 1 <= n; ++ j){
	st[j][i] = min(st[j][i-1], st[j+(1<<i-1)][i-1]);
	}
	}
	}
	int qry(int l, int r){
	if(l == r){
	return 1e9;
	}
	if(l > r){
	swap(l, r);
	}
	++ l;
	int k = 31 ^ __builtin_clz(r - l + 1);
	return min(st[l][k], st[r-(1<<k)+1][k]);
	}

	int t[NM], ls[NM], rs[N*M], rt[N], tot;
	int add(int p, int l, int r, int x){
	++ tot;
	tie(t[tot], ls[tot], rs[tot]) = tie(t[p], ls[p], rs[p]);
	p = tot;
	if(l == r){
	++ t[p];
	} else {
	int mid = l + r >> 1;
	if(x <= mid){
	ls[p] = add(ls[p], l, mid, x);
	} else {
	rs[p] = add(rs[p], mid+1, r, x);
	}
	t[p] = t[ls[p]] + t[rs[p]];
	}
	return p;
	}
	int sum(int p, int q, int l, int r, int ql, int qr){
	if(qr < l \|\| r < ql){
	return 0;
	} else if(ql <= l && r <= qr){
	return t[p] - t[q];
	} else {
	int mid = l + r >> 1;
	return sum(ls[p], ls[q], l, mid, ql, qr) + sum(rs[p], rs[q], mid+1, r, ql, qr);
	}
	}
	void seg(){
	for(int i = 1; i <= n; ++ i){
	rt[i] = add(rt[i-1], 1, n, rk[i]);
	}
	}

	bool chk(int a, int b, int c, int d, int k){
	int l = a, r = b - k + 1;
	int x, y;
	int L = 1, R = rk[c];
	while(L < R){
	int mid = L + R >> 1;
	if(qry(mid, rk[c]) >= k){
	R = mid;
	} else {
	L = mid + 1;
	}
	}
	x = L;
	L = rk[c], R = n;
	while(L < R){
	int mid = L + R + 1 >> 1;
	if(qry(rk[c], mid) >= k){
	L = mid;
	} else {
	R = mid - 1;
	}
	}
	y = L;
	return sum(rt[r], rt[l-1], 1, n, x, y);
	}

	void solve(){
	scanf("%d%d", &n, &q);
	scanf("%s", s + 1);
	suf();
	seg();
	table();
	while(q--){
	int a, b, c, d;
	scanf("%d%d%d%d", &a, &b, &c, &d);
	int l = 0, r = min(d - c + 1, b - a + 1);
	while(l < r){
	int mid = l + r + 1 >> 1;
	if(chk(a, b, c, d, mid)){
	l = mid;
	} else {
	r = mid - 1;
	}
	}
	printf("%d\n", l);
	}
	}

KiharaTouma

木原 当麻

2024.04 别急记录

1. 餐巾计划问题

2. 国家集训队 - happiness

3. qoj1427 - Flip

4. USACO 2024 OPEN - Cowreography G

5. CCO2019 - Sirtet

6. LuoguP3600 - 随机数生成器

7. AGC066A - Adjacent Difference

8. AGC066B - Decreasing Digit Sums

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11. NOI2023 - 贸易

12. THUPC 2024 初赛 - 多折较差验证

13. AGC035C - Skolem XOR Tree

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15. AGC030C - Coloring Torus

16. NOI2020 - 超现实树

17. NOI2020 - 制作菜品

18. LuoguP2633 - Count on a tree

19. 国家集训队 - middle

20. HEOI/TJOI2016 - 字符串

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