1 条题解
-
0
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