来自学长的馈赠3

A. 学数数

单调栈 + 排序 + 前缀和 + 二分

单调栈应该是左边乘右边，考场脑抽写成长度。

对$lowe\_bound$，$upper\_bound$使用不熟

code

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
inline int read(){
	int x = 0; char c; c = getchar();
	while(c < '0' || c > '9')c = getchar();
	while(c >= '0' && c <= '9')x = (x << 3 ) + (x << 1) + (c ^ 48), c = getchar();
	return x;
}
typedef long long ll;
const int maxn = 100005;
int sta[maxn], l[maxn], r[maxn], top, a[maxn], p[maxn];
ll len[maxn], s[maxn];
bool cmp(int x, int y){return a[x] < a[y];}
char c[3];
signed main(){
	int n = read(), Q = read();
	for(int i = 1; i <= n; ++i)a[i] = read();
	for(int i = 1; i <= n; ++i){
		while(top && a[sta[top]] < a[i])r[sta[top--]] = i - 1;
		l[i] = sta[top] + 1; sta[++top] = i;
	}
	while(top)r[sta[top--]] = n;
	for(int i = 1; i <= n; ++i)p[i] = i;
	sort(p + 1, p + n + 1, cmp);
	sort(a + 1, a + n + 1);
	for(int i = 1; i <= n; ++i)len[i] = (1ll * r[p[i]] - p[i] + 1) * (1ll * p[i] - l[p[i]] + 1);
	for(int i = 1; i <= n; ++i)s[i] = s[i - 1] + len[i];
	ll ans = 0;s[n + 1] = s[n];
	for(int i = 1; i <= Q; ++i){
		int x; scanf("%s%d",c, &x);
		if(c[0] == '>')ans = s[n] - s[upper_bound(a + 1, a + n + 1, x) - a - 1];
		if(c[0] == '<')ans = s[lower_bound(a + 1, a + n + 1, x) - a - 1];
		if(c[0] == '=')ans = s[upper_bound(a + 1, a + n + 1, x) - a - 1] - s[lower_bound(a + 1, a + n + 1, x) - a - 1];
		printf("%lld\n",ans);
	}
	return 0;
}

B. sum

容易得出$2 * S_n^m - C_n^m = S_{n + 1} ^ m$

这样再结合定义，我们可以在杨辉三角上$O(1)$上下左右移动，然后用莫队处理即可。？？

太奇妙了

code

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
inline int read(){
	int x = 0; char c; c = getchar();
	while(c < '0' || c > '9')c = getchar();
	while(c >= '0' && c <= '9')x = (x << 3 ) + (x << 1) + (c ^ 48), c = getchar();
	return x;
}
typedef long long ll;
const int maxn = 100000;
const int mod = 1e9 + 7;
const int len = 350;
int fac[maxn + 55], inv[maxn + 55];
inline int qpow(int x, int y){
	int ans = 1;
	for(; y; y >>= 1, x = 1ll * x * x % mod)if(y & 1)ans = 1ll * ans * x % mod;
	return ans;
}
inline void pre(){
	fac[0] = 1; for(register int i = 1; i <= maxn; ++i)fac[i] = 1ll * fac[i - 1] * i % mod;
	inv[maxn] = qpow(fac[maxn], mod - 2); for(register int i = maxn - 1; i; --i)inv[i] = 1ll * inv[i + 1] * (i + 1) % mod; inv[0] = 1;
}
inline int C(int n, int m){return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod;}
inline int S(int n, int m){ll ans = 0;for(register int i = 0; i <= m; ++i)ans += C(n, i);return ans % mod;}
int block[maxn + 55], nn, nm, ans[maxn + 55]; ll nans;
struct query{int n, m, id;}q[maxn + 55];
bool cmp(query x, query y){return block[x.n] == block[y.n] ? (block[x.n] & 1 ? x.m < y.m : x.m > y.m) : x.n < y.n;}
inline void add_n(){nans = (nans + nans - C(nn, nm) + mod) % mod;}
inline void red_n(){nans = (nans + C(nn - 1,nm)) * inv[2] % mod;}
inline void add_m(){nans = (nans + C(nn, nm + 1)) % mod;}
inline void red_m(){nans = (nans - C(nn, nm) + mod) % mod;}
inline void work(int Q){
	ans[q[1].id] = S(q[1].n, q[1].m);
	nn = q[1].n, nm = q[1].m, nans = ans[q[1].id];
	for(register int i = 2; i <= Q; ++i){
		for(; nn < q[i].n; ++nn)add_n();
		for(; nm > q[i].m; --nm)red_m();
		for(; nn > q[i].n; --nn)red_n();
		for(; nm < q[i].m; ++nm)add_m();
		ans[q[i].id] = nans;
	}
}
int main(){
	pre();
	int id = read(), Q = read();
	for(register int i = 1; i <= Q; ++i)q[i].n = read(), q[i].m = read();
	for(register int i = 1; i <= Q; ++i)q[i].id = i;
	for(register int i = 1; i <= maxn; ++i)block[i] = (i + len - 1) / len;
	sort(q + 1, q + Q + 1, cmp);
	work(Q);
	for(register int i = 1; i <= Q; ++i)printf("%d\n",ans[i]);
	return 0;
}

C. 完美子图

两种解法

1. 分治

2. 线段树

首先读入$x,y$时令$a[x] = y$问题转化为多少区间值域连续，即$max - min = r - l$

先说分治

$solve(l, r)$处理当前区间

$l - mid$，$mid + 1 - r$递归处理

考虑如何处理两侧合并的贡献

考虑两种情况，即最大值最小值在一侧/两侧

在一侧时根据$max - min = r - l$知三求一，判一下是否合法即可

在两侧时，以左小右大为例

移项得$l - min = r - max$，

我们从$mid - > l$枚举$i$，右区间进行双指针

实际上因为$max_i,min_i$的存在右区间因为$max$有一个下界，根据$min$有一个上界，而且这两个界一定是单调右移的，我们把符合要求的加进桶里，左边直接查桶即可

code

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
inline int read(){
	int x = 0; char c; c = getchar();
	while(c < '0' || c > '9')c = getchar();
	while(c >= '0' && c <= '9')x = (x << 3 ) + (x << 1) + (c ^ 48), c = getchar();
	return x;
}
typedef long long ll;
const int maxn = 500005;
int n, a[maxn], mi[maxn], mx[maxn], cnt[maxn << 1 | 1];
ll ans = 0;
void solve(int l, int r){
	if(l == r){++ans;return;}
	int mid = (l + r) >> 1;
	solve(l, mid); solve(mid + 1, r);
	mx[mid] = mi[mid] = a[mid];
	mx[mid + 1] = mi[mid + 1] = a[mid + 1];
	for(int i = mid - 1; i >= l; --i){
		mx[i] = max(mx[i + 1], a[i]);
		mi[i] = min(mi[i + 1], a[i]);
	}
	for(int i = mid + 2; i <= r; ++i){
		mx[i] = max(mx[i - 1], a[i]);
		mi[i] = min(mi[i - 1], a[i]);
	}
	for(int i = mid; i >= l; --i){
		int j = mx[i] - mi[i] + i;
		if(j > mid && j <= r && mi[j] > mi[i] && mx[j] < mx[i]) ++ans;
	}
	for(int i = mid + 1; i <= r; ++i){
		int j = i - mx[i] + mi[i];
		if(j <= mid && j >= l && mi[j] > mi[i] && mx[j] < mx[i]) ++ans;
	}
	int j = mid + 1, k = mid + 1;
	for(int i = mid; i >= l; --i){
		while(j <= r && mi[j] > mi[i]){++cnt[mx[j] - j + n]; ++j;}
		while(k < j && mx[k] < mx[i]){--cnt[mx[k] - k + n];++k;}
		ans += cnt[mi[i] - i + n];
	}
	while(k < j){--cnt[mx[k] - k + n]; ++k;}
	j = mid + 1, k = mid + 1;
	for(int i = mid; i >= l; --i){
		while(j <= r && mx[j] < mx[i]){++cnt[mi[j] + j]; ++j;}
		while(k < j && mi[k] > mi[i]){--cnt[mi[k] + k]; ++k;}
		ans += cnt[mx[i] + i];
	}
	while(k < j){--cnt[mi[k] + k]; ++k;}
	// j = mid, k = mid;
	// for(int i = mid + 1; i <= r; ++i){
	// 	while(j >= l && mi[j] > mi[i]){++cnt[mx[j] + j]; --j;}
	// 	while(k > j && mx[k] < mx[i]){--cnt[mx[k] + k];--k;}
	// 	ans += cnt[mi[i] + i];
	// }
	// while(k > j){--cnt[mx[k] + k]; --k;}
}
int main(){
	n = read();
	for(int i = 1; i <= n; ++i){int x = read(), y = read();a[x] = y;}
	solve(1, n);
	printf("%lld\n",ans);
	return 0;
}

线段树做法

从左往右第$i$个节点维护第$i - r$的答案

根据$max- min + l >= r$我们动态修改后查询最小值以及最小值个数即可统计答案

更新哪个区间?单调栈的同时进行更新即可

D. DP搬运工2

用之前的式子改了下转移，水到$80$

其实之前的式子中的$j$在这里可以砍掉

转移的话考虑当前数插在两边/两边的数原来有无贡献

code

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;
typedef long long ll;
const int maxn = 2051;
const int mod = 998244353;
int n, K;
ll f[maxn][maxn];
int main(){
	scanf("%d%d", &n, &K);
	f[1][0] = 1;
	for(int i = 1; i < n; ++i){
		for(int k = 0; k <= K; ++k)
		if(f[i][k]){
			f[i][k] %= mod;
			f[i + 1][k] += 2 * (k + 1) * f[i][k];
			f[i + 1][k + 1] += (i - 1 - k * 2) * f[i][k];
		}
	}
	printf("%lld\n",f[n][K] % mod);
	return 0;
}