全加和 UVa10943

来源：互联网发布：舆情分析研判数据编辑：程序博客网时间：2024/05/06 01:19

1.题目描述：点击打开链接

2.解题思路：本题实际上就是求x1+x2+...+xk=n的非负整数解的个数。根据组合数学的结论，答案是C(n+k-1,k-1)。可以事先预处理算出所有的组合数。

3.代码：

#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair<int, int> P;typedef pair<long long, long long> PL;#define me(s) memset(s,0,sizeof(s))#define For(i,n) for(int i=0;i<(n);i++)const int MOD = 1000000;int n, k;#define N 205ll C[N][N];void init(){me(C);C[0][0] = 1;for (int i = 1; i < 200; i++){C[i][0] = C[i][i] = 1;for (int j = 1; j < i; j++)C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;}}int main(){//freopen("t.txt", "r", stdin);init();while (~scanf("%d%d", &n, &k)){if (!n&&!k)break;int ans = C[n + k - 1][k - 1];printf("%d\n", ans);}return 0;}

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