UVa10943
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简介:把K个不超过N的非负整数加起来,使ta们的和为N
分析:相当于有N个小球,要放到K个盒子里,允许有空盒子,问方案数
ans=C(N+K-1,K-1)
tip
读入的时候如果不写!=EOF,读入就不会正常结束,对拍的时候就拍不起来
//这里写代码片#include<cstdio>#include<cstring>#include<iostream>using namespace std;const int mod=1000000;int C[202][202];void cl(){ int i,j; C[0][0]=1; C[1][1]=1; C[1][0]=1; for (i=2;i<=200;i++) { C[i][0]=1; C[i][i]=1; for (j=1;j<i;j++) C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod; }}int main(){ int n,k; cl(); while (scanf("%d%d",&n,&k)!=EOF) { printf("%d\n",C[n+k-1][k-1]); } return 0;}
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