python/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_2.cpp

#include <bits/stdc++.h>
/*
测试用例：
5 6
1 0 1 0 0 0
0 1 0 1 0 0
1 0 0 0 1 0
0 1 0 0 0 1
1 0 1 0 0 0


测试用例II：
5 6
1 0 1 0 0 0         2 0 4 0 0 0
0 1 0 1 1 0         0 1 0 3 1 0
1 0 0 0 1 0         3 0 0 0 2 0
0 1 0 0 0 1         0 2 0 0 0 1
1 0 1 0 0 0         1 0 1 0 0 0

5 6
1 0 1 0 0 0
0 1 0 1 1 0
1 0 0 0 1 0
0 1 0 0 0 1
1 0 1 0 0 0


测试用例III：
5 6
1 0 1 0 0 0
0 1 0 1 1 0
1 0 0 0 1 0
0 1 0 1 0 1
1 0 1 0 1 0

*/
using namespace std;
const int N = 110;
int a[N][N];
int res;

int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> a[i][j];

    //直接在原数组上递推
    for (int i = n; i >= 1; i--)            //从倒数第二行开始
        for (int j = 1; j <= m; j++)        //这个需要考虑一下边界问题
            if (a[i][j])                    //只有当前位置大于0才有考虑的价值
                a[i][j] += a[i + 1][j + 1]; //递推式

    cout << "====================" << endl;
    //打印出这个新的递推数组看看是什么情况
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++)
            cout << a[i][j] << " ";
        cout << endl;
    }
    return 0;
}