https://ac.nowcoder.com/acm/contest/332/E

C++版本一

题解：二维前缀和DP

dp[i][j]代表从（1,1）到（i，j）的所有低于d的数量

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,q,x,y,x2,y2,d;
int ans,cnt,flag,temp;
char str;

int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    scanf("%d%d%d",&n,&m,&d);
    int a[n+10][m+10];
    int dp[n+10][m+10];
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            scanf("%d",&a[i][j]);
            if(a[i][j]>=d){
                dp[i][j]++;
            }
            dp[i][j]=dp[i][j]+dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1];
        }
    }
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d%d",&x,&y,&x2,&y2);
        cout << dp[x2][y2]-dp[x-1][y2]-dp[x2][y-1]+dp[x-1][y-1]<< endl;
    }
    //cout << "Hello world!" << endl;
    return 0;
}