求 组合数 dp

 1 从组合数的定义可以知道。C(n, m) = C(n-1, m) + C(n-1, m-1);
 2 所以,根据这个公式可以递归的求得组合数,代码如下:
 3 
 4 long long C(int n, int m) {
 5     if(m == 0 || n == 0 || n == 1 || m == n)     return 1;
 6     if(aug[n][m] != 0)     return aug[n][m];
 7     
 8     aug[n-1][m] = C(n-1, m);
 9     aug[n-1][m-1] = C(n-1, m-1);
10 
11     return aug[n-1][m] + aug[n-1][m-1];
12 }
 1 非递归的写法:
 2 
 3 long long C(int n, int m) {
 4     int i, j;
 5     for(i = 0; i <= m; i++)
 6         aug[0][i] = aug[1][i] = 1;
 7     for(i = 0; i <= n; i++)
 8         aug[i][0] = 1;
 9     for(i = 0; i <= m; i++) {aug[i][i] = 1;}
10     for(i = 1; i <= n; i++) {
11         for(j = 1; j <= m; j++) {
12             if(i != j)
13                 aug[i][j] = aug[i-1][j] + aug[i-1][j-1];
14         }
15     }
16     return aug[n][m];
17 }
原文地址:https://www.cnblogs.com/acSzz/p/2519371.html