• 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 << "
    ";
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jungu/p/13362537.html
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