2020.7华东月赛

思路：

先求出二维前缀和，再用差分枚举每个点的值

Code：

#pragma GCC optimize(3)
#pragma GCC optimize(2)
#include <map>
#include <set>
#include <array>
#include <queue>
#include <stack>
#include <cmath>
#include <vector>
#include <cstdio>
#include <cstring>
#include <sstream>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
#include <unordered_map>

using namespace std;

typedef long long ll;
typedef pair<int, int> PII;

#define Time (double)clock() / CLOCKS_PER_SEC

#define sd(a) scanf("%d", &a)
#define sdd(a, b) scanf("%d%d", &a, &b)
#define slld(a) scanf("%lld", &a)
#define slldd(a, b) scanf("%lld%lld", &a, &b)

const int N = 5e3 + 10; //4e3 + 10;
const ll M = 4e12;
const int mod = 1e9 + 7;

int n, m, p, q, maxx;
int a[N][N], sum[N][N], ans[N][N];

void update(int x, int y)
{
    for (int i = 1; i <= p; i++)
    {
        for (int j = 1; j <= q; j++)
        {
            
            ans[x + i - 1][y + j - 1] += a[i][j];
            // cout << x + i - 1 << " " << y - 1 + j << " " << ans[x + i - 1][y + j - 1] << endl; 
            maxx = max(maxx, ans[x + i - 1][y + j - 1]);
        }
    }
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("/home/jungu/code/in.txt", "r", stdin);
    // freopen("/home/jungu/code/out.txt", "w", stdout);
#endif
    // ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    sdd(p, q);
    sdd(n, m);

    for (int i = 1; i <= p; i++)
    {
        for (int j = 1; j <= q; j++)
        {
            sd(a[i][j]);
            sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + a[i][j];
        }
    }
    for(int i = 1; i <= n; i ++){
        for(int j = 1; j <= m; j ++){
            int x1 = min(i, p);
            int y1 = min(j, q);
            int x2 = max(0, i - (n - p + 1)) + 1;
            int y2 = max(0, j - (m - q + 1)) + 1;
            ans[i][j] = sum[x1][y1] - sum[x2 - 1][y1] - sum[x1][y2 - 1] + sum[x2 - 1][y2 - 1];
            maxx = max(maxx, ans[i][j]);
        }
    }
    

    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            ans[i][j] = ans[i][j] * 100 / maxx;
            cout << ans[i][j] << " ";
        }
        cout << "\n";
    }

    return 0;
}