「PKUWC2018/PKUSC2018」试题选做
「PKUWC2018/PKUSC2018」试题选做
最近还没想好报THUSC还是PKUSC，THU发我的三类约（再来一瓶）不知道要不要用，甚至不知道营还办不办，协议还有没有用。所以这些事情就暂时先不管了，PKU的题还是不错的，就刷一刷划水。因为比较简单，所以就不单独写博客了。
loj2537 Minimax
数据结构题，两个 \(\log\) 直接启发式合并，一个 \(\log\) 需要转移的时候多维护一些东西。
对于每个节点维护一下选择其子树里每个叶子的权值的概率，线段树合并转移即可。
code
/*program by mangoyang*/
#include <bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int M = 1e9, N = 300005, mod = 998244353;
int a[N], rt[N], P, n, ans;
vector<int> g[N];
vector<pair<int, int> > Ans;
namespace Seg{
#define mid ((l + r) >> 1)
int s[N*30], tag[N*30], lc[N*30], rc[N*30], size;
inline void init(){ for(int i = 1; i < N * 30; i++) tag[i] = 1; }
inline void pushdown(int u){
if(tag[u] == 1) return;
if(lc[u]){
s[lc[u]] = 1ll * tag[u] * s[lc[u]] % mod;
tag[lc[u]] = 1ll * tag[lc[u]] * tag[u] % mod;
}
if(rc[u]){
s[rc[u]] = 1ll * tag[u] * s[rc[u]] % mod;
tag[rc[u]] = 1ll * tag[rc[u]] * tag[u] % mod;
}
tag[u] = 1;
}
inline void insert(int &u, int l, int r, int pos){
if(!u) u = ++size;
if(l == r) return (void) (s[u] = 1);
if(pos <= mid) insert(lc[u], l, mid, pos);
else insert(rc[u], mid + 1, r, pos);
s[u] = s[lc[u]] + s[rc[u]];
}
inline int merge(int x, int y, int sumx, int sumy){
if(!x){
if(!sumx) return y;
s[y] = 1ll * s[y] * sumx % mod;
tag[y] = 1ll * tag[y] * sumx % mod;
return y;
}
if(!y){
if(!sumy) return x;
s[x] = 1ll * s[x] * sumy % mod;
tag[x] = 1ll * tag[x] * sumy % mod;
return x;
}
pushdown(x), pushdown(y);
int o = ++size;
int Lx = (sumx + 1ll * P * s[lc[x]] % mod) % mod;
int Ly = (sumy + 1ll * P * s[lc[y]] % mod) % mod;
int Rx = (sumx + 1ll * (mod + 1 - P) * s[rc[x]] % mod) % mod;
int Ry = (sumy + 1ll * (mod + 1 - P) * s[rc[y]] % mod) % mod;
lc[o] = merge(lc[x], lc[y], Rx, Ry);
rc[o] = merge(rc[x], rc[y], Lx, Ly);
s[o] = (s[lc[o]] + s[rc[o]]) % mod;
return o;
}
inline void getans(int u, int l, int r){
if(l == r) return (void) Ans.push_back(make_pair(l, s[u]));
pushdown(u);
if(lc[u]) getans(lc[u], l, mid);
if(rc[u]) getans(rc[u], mid + 1, r);
}
}
inline int Pow(int a, int b){
int ans = 1;
for(; b; b >>= 1, a = 1ll * a * a % mod)
if(b & 1) ans = 1ll * ans * a % mod;
return ans;
}
inline void dfs(int u){
if(!(int) g[u].size()) Seg::insert(rt[u], 1, M, a[u]);
for(int i = 0; i < (int) g[u].size(); i++){
dfs(g[u][i]);
P = a[u];
rt[u] = Seg::merge(rt[u], rt[g[u][i]], 0, 0);
}
}
int main(){
read(n);
for(int i = 1, x; i <= n; i++) read(x), g[x].push_back(i);
for(int i = 1; i <= n; i++){
read(a[i]);
if((int) g[i].size()) a[i] = 1ll * a[i] * Pow(10000, mod1- 2) % mod;
}
Seg::init();
dfs(1), Seg::getans(rt[1], 1, M), sort(Ans.begin(), Ans.end());
for(int i = 0; i < (int) Ans.size(); i++){
(ans += 1ll * (i + 1) * Ans[i].first % mod * Ans[i].second % mod * Ans[i].second % mod) %= mod;
}
cout << ans << endl;
return 0;
}
loj2538 Slay the Spire
一个显然的贪心策略是从大到小尽可能打出强化牌（至多 \(k-1\) 张），然后从大到小尽可能打出攻击牌。
可以分别对两种牌排序然后按照这个策略设计 dp。
\(f[i][j]\) 表示前 \(i\) 张强化牌摸了 \(j\) 张的期望加成，从大到小对强化牌排序，枚举当前这张牌是否被摸到来转移。
\(g[i][j]\) 表示前 \(i\) 张攻击牌摸了 \(j\) 张的期望攻击力之和，\(j\) 越大可以打的攻击牌越多，所以从小到大排序来转移。
code
/*program by mangoyang*/
#include <bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int N = 3005, mod = 998244353;
int a[N], b[N], js[N], inv[N], f[2][N], g[2][N], n, m, k;
inline void up(int &x, int y){
x = (x + y >= mod ? x + y - mod : x + y);
}
inline int Pow(int a, int b){
int ans = 1;
for(; b; b >>= 1, a = 1ll * a * a % mod)
if(b & 1) ans = 1ll * ans * a % mod;
return ans;
}
inline int C(int x, int y){
return 1ll * js[x] * inv[y] % mod * inv[x-y] % mod;
}
inline void solve(){
read(n), read(m), read(k);
for(int i = 1; i <= n; i++) read(a[i]);
for(int i = 1; i <= n; i++) read(b[i]);
sort(a + 1, a + n + 1, greater<int>());
sort(b + 1, b + n + 1);
for(int i = 0; i <= n; i++)
f[0][i] = f[1][i] = g[0][i] = g[1][i] = 0;
f[0][0] = 1;
for(int i = 1, o = 1; i <= n; i++, o ^= 1){
for(int j = 0; j <= i; j++){
f[o][j] = f[o^1][j];
if(j && j < k) up(f[o][j], 1ll * a[i] * f[o^1][j-1] % mod);
else if(j) up(f[o][j], f[o^1][j-1]);
}
}
for(int i = 1, o = 1; i <= n; i++, o ^= 1){
for(int j = 0; j <= i; j++){
g[o][j] = g[o^1][j];
if(j){
if(m - j < k - 1) up(g[o][j], g[o^1][j-1]);
up(g[o][j], 1ll * C(i - 1, j - 1) * b[i] % mod);
}
}
}
int ans = 0;
for(int i = 0; i <= min(m, n); i++)
if(m - i <= n) up(ans, 1ll * f[n&1][i] * g[n&1][m-i] % mod);
printf("%d\n", ans);
}
int main(){
js[0] = inv[0] = 1;
for(int i = 1; i < N; i++)
js[i] = 1ll * js[i-1] * i % mod, inv[i] = Pow(js[i], mod - 2);
int T; read(T); while(T--) solve();
return 0;
}
loj2540 随机算法
直接状压 \(dp\) 是 \(O(3^nn)\) 的，实际上选了会冲突的点与已经选了的点显然可以看做同一类点，然后就 \(O(2^nn)\) 了。
code
/*program by mangoyang*/
#include<bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int mod = 998244353;
int cnt[1<<21], dp[1<<21][21], js[21], inv[21], ed[21], n, m;
inline int Pow(int a, int b){
int ans = 1;
for(; b; b >>= 1, a = 1ll * a * a % mod)
if(b & 1) ans = 1ll * ans * a % mod;
return ans;
}
inline int A(int x, int y){
return 1ll * js[x] * inv[x-y] % mod;
}
inline void up(int &x, int y){
x = (x + y >= mod ? x + y - mod : x + y);
}
int main(){
read(n), read(m);
js[0] = inv[0] = 1;
for(int i = 1; i <= n; i++){
js[i] = 1ll * js[i-1] * i % mod;
inv[i] = Pow(js[i], mod - 2);
}
for(int i = 1, x, y; i <= m; i++){
read(x), read(y);
ed[x] |= 1 << (y - 1);
ed[y] |= 1 << (x - 1);
}
for(int s = 0; s < (1 << n); s++)
for(int i = 0; i < n; i++)
if((1 << i) & s) cnt[s]++;
dp[0][0] = 1;
for(int s = 0; s < (1 << n) - 1; s++)
for(int i = 0; i < n; i++) if(dp[s][i]){
for(int j = 0; j < n; j++) if(!((1 << j) & s)){
int t = s | (1 << j) | ed[j+1];
up(dp[t][i+1], 1ll * dp[s][i] * A(n - cnt[s] - 1, cnt[s^t] - 1) % mod);
}
}
int mx = 0;
for(int i = 1; i <= n; i++) if(dp[(1<<n)-1][i]) mx = i;
cout << 1ll * dp[(1<<n)-1][mx] * inv[n] % mod << endl;
return 0;
}
loj2542 随机游走
设 \(dp[x][s]\) 为从 \(x\) 出发走到点集 \(s\) 中任意一个点的期望步数，然后 \(\min-\max\) 容斥即可 ，直接分层图+高斯消元跑不过去，可以用小套路把未知数设出来以后每个点都可以用其父亲的答案乘上一个系数再加上一个常数来表示，然后跑一遍树形dp就做完了。
code
/*program by mangoyang*/
#include<bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int N = 1 << 21, mod = 998244353;
vector<int> g[N];
int mn[N], cnt[N], in[N], A[N], B[N], n, q, rt;
inline int Pow(int a, int b){
int ans = 1;
for(; b; b >>= 1, a = 1ll * a * a % mod)
if(b & 1) ans = 1ll * ans * a % mod;
return ans;
}
inline void dfs(int u, int fa){
if(in[u]) return (void) (A[u] = B[u] = 0);
int SA = 0, SB = 0, deg = (int) g[u].size();
for(int i = 0; i < (int) g[u].size(); i++){
int v = g[u][i];
if(v == fa) continue;
dfs(v, u);
SA = (SA + A[v]) % mod, SB = (SB + B[v]) % mod;
}
deg = Pow(deg, mod - 2);
int x = (1 - 1ll * SA * deg % mod + mod) % mod;
x = Pow(x, mod - 2);
A[u] = 1ll * deg * x % mod;
B[u] = 1ll * x * (1ll * SB * deg % mod + 1) % mod;
}
int main(){
read(n), read(q), read(rt);
for(int i = 1, x, y; i < n; i++){
read(x), read(y);
g[x].push_back(y), g[y].push_back(x);
}
for(int s = 0; s < (1 << n); s++){
for(int i = 1; i <= n; i++){
in[i] = ((1 << (i - 1)) & s) ? 1 : 0;
cnt[s] += in[i];
}
dfs(rt, 0), mn[s] = B[rt];
}
while(q--){
int num, sta = 0;
read(num);
for(int i = 1, x; i <= num; i++) read(x), sta |= 1 << (x - 1);
int ans = 0;
for(int s = sta; s; s = (s - 1) & sta)
(ans += 1ll * ((cnt[s] & 1) ? 1 : mod - 1) * mn[s] % mod) %= mod;
printf("%d\n", ans);
}
return 0;
}
loj6432 真实排名
对于每个点枚举一下要不要，统计一下会影响的人数，组合数算一下就好了
code
/*program by mangoyang*/
#include <bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
#define fi first
#define se second
const int N = 1000005, mod = 998244353;
pair<int, int> a[N];
int js[N], inv[N], ans[N], n, k;
inline int Pow(int a, int b){
int ans = 1;
for(; b; b >>= 1, a = 1ll * a * a % mod)
if(b & 1) ans = 1ll * ans * a % mod;
return ans;
}
inline int C(int x, int y){
if(x < y || x < 0 || y < 0) return 0;
return 1ll * js[x] * inv[y] % mod * inv[x-y] % mod;
}
int main(){
read(n), read(k);
js[0] = inv[0] = 1;
for(int i = 1; i <= n; i++){
js[i] = 1ll * js[i-1] * i % mod;
inv[i] = Pow(js[i], mod - 2);
read(a[i].fi), a[i].se = i;
}
sort(a + 1, a + n + 1);
int p = 0, q = 0;
for(int i = 1; i <= n; i++){
if(a[i].fi == a[i-1].fi && i > 1){
ans[a[i].se] = ans[a[i-1].se];
continue;
}
while(a[p+1].fi * 2 < a[i].fi && p < n) p++;
while(a[q+1].fi < a[i].fi * 2 && q < n) q++;
int tot1 = i - p - 1, tot2 = max(q, i) - i + 1;
(ans[a[i].se] += C(n - tot1 - 1, k)) %= mod;
(ans[a[i].se] += C(n - tot2, k - tot2)) %= mod;
}
for(int i = 1; i <= n; i++) printf("%d\n", ans[i]);
return 0;
}
loj6433 最大前缀和
考虑每一个集合成为最大前缀和的方案数，这个等于其内部排列满足全部选是最大前缀和乘上剩余元素无论怎么排都不存在一个前缀和 \(>0\) 的方案数，两部分分开来状压 \(dp\) 即可。
code
/*program by mangoyang*/
#include <bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int N = (1 << 20), mod = 998244353;
int a[N], f[N], sum[N], g[N], n;
inline void up(int &x, int y){
x = x + y >= mod ? x + y - mod : x + y;
}
int main(){
read(n);
for(int i = 1; i <= n; i++) read(a[i]);
for(int s = 1; s < (1 << n); s++)
for(int i = 0; i < n; i++)
if((1 << i) & s) sum[s] += a[i+1];
g[0] = 1;
for(int i = 0; i < n; i++) f[1<<i] = 1;
for(int s = 0; s < (1 << n); s++)
for(int i = 0; i < n; i++) if(!((1 << i) & s)){
int t = s ^ (1 << i);
if(sum[s] > 0) up(f[t], f[s]);
if(sum[t] <= 0) up(g[t], g[s]);
}
int ans = 0;
for(int s = 0; s < (1 << n); s++){
sum[s] = (sum[s] % mod + mod) % mod;
up(ans, 1ll * sum[s] * f[s] % mod * g[((1<<n)-1)^s] % mod);
}
cout << ans << endl;
return 0;
}
loj6436 星际穿越
可以发现每个点出发最多会向右走一步，于是倍增出每个点走 \(2^j\) 步能走到的最左端点，已经走到这之间所有点的距离和，每次询问拆成两个以 \(x\) 结尾的后缀来询问即可。
code
/*program by mangoyang*/
#include <bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int N = 300005;
ll a[N], f[N][21], s[N][21];
int n, q;
inline ll calc(int x, int L){
if(a[x] <= L) return x - L;
ll ans = x - a[x], now = 1; x = a[x];
for(int i = 20; ~i; i--) if(f[x][i] >= L){
ans += s[x][i] + (x - f[x][i]) * now;
now += 1 << i, x = f[x][i];
}
return ans + (x - L) * (now + 1);
}
int main(){
read(n), a[1] = 1;
for(int i = 2; i <= n; i++) read(a[i]);
for(int i = n; i >= 1; i--){
f[i][0] = min(a[i], i == n ? inf : f[i+1][0]);
s[i][0] = i - f[i][0];
}
for(int j = 1; j <= 20; j++)
for(int i = 1; i <= n; i++){
f[i][j] = f[f[i][j-1]][j-1];
s[i][j] = s[i][j-1] + s[f[i][j-1]][j-1];
s[i][j] += (f[i][j-1] - f[f[i][j-1]][j-1]) << (j - 1);
}
read(q);
for(int i = 1, l, r, x; i <= q; i++){
read(l), read(r), read(x);
ll ans = calc(x, l) - calc(x, r + 1);
ll g = __gcd(ans, (ll) r - l + 1);
printf("%lld/%lld\n", ans / g, (r - l + 1) / g);
}
return 0;
}
loj6436 神仙的游戏
border有关的计数题可以从等差数列和周期两个角度考虑，这个题显然不能等差数列于是就直接上周期定理。
套用周期定理得到如果 \(i\) 是一个 border，那么所有模 \(|S|-i\) 同余的位置必须相同，也就是不能出现两个距离为 \(|S|-i-1\) 的 \(0,1\) 位置，直接用 FTT 统计所有差的信息即可。
code
/*program by mangoyang*/
#include <bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int N = (1 << 22), P = 998244353, G = 3;
char s[N];
int a[N], b[N];
namespace poly{
int rev[N], lg, len;
inline int Pow(int a, int b){
int ans = 1;
for(; b; b >>= 1, a = 1ll * a * a % P)
if(b & 1) ans = 1ll * ans * a % P;
return ans;
}
inline void timesinit(int lenth){
for(len = 1, lg = 0; len <= lenth; len <<= 1, lg++);
for(int i = 0; i < len; i++)
rev[i] = (rev[i>>1] >> 1) | ((i & 1) << (lg - 1));
}
inline void dft(int *a, int sgn){
for(int i = 0; i < len; i++)
if(i < rev[i]) swap(a[i], a[rev[i]]);
for(int k = 2; k <= len; k <<= 1){
int w = Pow(G, (P - 1) / k);
if(sgn == - 1) w = Pow(w, P - 2);
for(int i = 0; i < len; i += k){
int now = 1;
for(int j = i; j < i + (k >> 1); j++){
int x = a[j], y = 1ll * a[j+(k>>1)] * now % P;
a[j] = x + y >= P ? x + y - P : x + y;
a[j+(k>>1)] = x - y < 0 ? x - y + P : x - y;
now = 1ll * now * w % P;
}
}
}
if(sgn == -1){
int Inv = Pow(len, P - 2);
for(int i = 0; i < len; i++)
a[i] = 1ll * a[i] * Inv % P;
}
}
}
using poly::timesinit;
using poly::dft;
int main(){
scanf("%s", s); int n = strlen(s);
for(int i = 0; i < n; i++)
a[i] = s[i] == '0', b[i] = s[i] == '1';
reverse(b, b + n);
timesinit(n + n - 1);
dft(a, 1), dft(b, 1);
for(int i = 0; i < poly::len; i++)
a[i] = 1ll * a[i] * b[i] % P;
dft(a, -1);
ll ans = 1ll * n * n;
for(int i = 1; i < n; i++){
int flag = 0;
for(int j = i; j < n; j += i)
if(a[n-1+j] || a[n-1-j]) flag = 1;
if(!flag) ans ^= 1ll * (n - i) * (n - i);
}
cout << ans << endl;
return 0;
}
