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和入阵曲那题很像

这里 \(n\) 很小，可以直接 \(n^2\) 压成一维考虑
然后就是对每个 \(j\) 查询 \([j-r, j-l]\) 中数的个数
这里我是用树状数组求的，带个log，被卡成了80pts
发现随着 \(j\) 单增， \(j-r, j-l\) 单调不减
所以可以双指针
题目里这些奇奇怪怪的单调性如何发现啊……

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long 
#define reg register int
//#define int long long 

inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, m, L, R;
int mat[35][50010], sum[35][50010];
char tem[50010];

namespace force{
	ll ans;
	void solve() {
		for (int i=1; i<=n; ++i) {
			for (int j=1; j<=m; ++j) 
				sum[i][j] = sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+mat[i][j];
		}
		for (int i=1; i<=n; ++i) {
			for (int j=1; j<=m; ++j) {
				for (int k=1; k<=i; ++k) {
					for (int h=1; h<=j; ++h) {
						int t=sum[i][j]-sum[k-1][j]-sum[i][h-1]+sum[k-1][h-1];
						if (t>=L && t<=R) ++ans;
					}
				}
			}
		}
		printf("%lld\n", ans);
		exit(0);
	}
}

namespace task1{
	ll ans;
	int a[1500010], lim;
	inline void upd(int i) {for (; i<=lim; i+=i&-i) ++a[i];}
	inline int query(int i) {int ans=0; for (; i; i-=i&-i) ans+=a[i]; return ans;}
	void solve() {
		lim=n*m+5;
		for (reg j=1; j<=m; ++j) 
			for (reg i=1; i<=n; ++i) 
				sum[i][j] = sum[i-1][j]+mat[i][j];
		for (reg l=1,sum2; l<=n; ++l) {
			for (reg i=l; i<=n; ++i) {
				//cout<<i<<": "<<l<<endl;
				memset(a, 0, sizeof(int)*lim);
				sum2=1; upd(1);
				for (reg j=1; j<=m; ++j) {
					sum2 += sum[i][j]-sum[i-l][j];
					if (sum2>L) ans+=query(sum2-L)-(sum2-R>1?query(max(sum2-R-1, 0)):0);
					upd(sum2);
				}
			}
		}
		printf("%lld\n", ans);
		exit(0);
	}
}

namespace task{
	ll ans;
	int cnt[1500010], cnt2[1500010];
	void solve() {
		for (reg j=1; j<=m; ++j) 
			for (reg i=1; i<=n; ++i) 
				sum[i][j] = sum[i-1][j]+mat[i][j];
		for (reg l=1,pos1,pos2,sum2; l<=n; ++l) {
			for (reg i=l; i<=n; ++i) {
				sum2=1; pos1=pos2=0; cnt[1]=1;
				for (reg j=1; j<=m; ++j) {
					sum2 += sum[i][j]-sum[i-l][j];
					while (pos1<sum2-R-1) {cnt[pos1]=cnt2[pos1]=0; ++pos1;}
					while (pos2<sum2-L) {++pos2; cnt2[pos2]=cnt2[pos2-1]+cnt[pos2];}
					if (sum2>L) ans+=cnt2[pos2]-cnt2[pos1];
					++cnt[sum2];
					if (pos2==sum2) ++cnt2[pos2];
				}
				for (reg j=pos1; j<=sum2; ++j) cnt[j]=cnt2[j]=0;
			}
		}
		printf("%lld\n", ans);
		exit(0);
	}
}

signed main()
{
	n=read(); m=read();
	for (reg i=1; i<=n; ++i) {
		scanf("%s", tem+1);
		for (reg j=1; j<=m; ++j)
			mat[i][j]=tem[j]-'0';
	}
	L=read(); R=read();
	task::solve();
	
	return 0;
}