1 条题解

  • 0
    @ 2025-11-6 12:06:03

    按题意直接模拟即可。

    AC code:

    #include<bits/stdc++.h>
    using namespace std;
    int n, m;
    long long dp[105][105];
    const int mod = 1e9 + 7;
    
    int main() {
        cin >> n >> m;
        for (int i = 1; i <= n; i++) {
            dp[i][1] = 1;
        }
        for (int i = 1; i <= m; i++) {
            dp[1][i] = 1;
        }
        for (int i = 2; i <= n; i++) {
            for (int j = 2; j <= m; j++) {
                dp[i][j] = (dp[i-1][j] + dp[i][j-1]) % mod;
            }
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                cout << dp[i][j] << " ";
            }
            cout << endl;
        }
        return 0;
    }
    
    • 1

    信息

    ID
    138
    时间
    1000ms
    内存
    128MiB
    难度
    1
    标签
    递交数
    3
    已通过
    3
    上传者