动态规划学习一

//C(n , k)的计算问题表述为了（问题描述）C(n-1 , k -1)和C(n -1, k)

C(n , k)=C(n-1, k -1)+C(n -1, k)

#include <cstdio>#include <iostream>using namespace std;int main(){int a[100][100];int index,n,i,j,n1;scanf("%d",&index);while (index-- ){scanf("%d%d",&n,&n1);memset(a , 0 , sizeof(a));for ( i = 1 ; i <= n ;++i ){a[i][1] = i ;a[i][i] = 1 ;}for ( i = 2 ; i <= n ; ++i ){for (j = 2; j <= n1; ++j ){a[i][j] = a[i-1][j-1] + a[i-1][j];}}printf("%d\n",a[n][n1]);}return 0;}

思路是从a[0][0]开始向后进行逐一的叠加