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XidianOJ 1183 Water Problem: Items divided

时间：2016-12-10 22:40:26 阅读：310 评论：0 收藏：0 [点我收藏+]

标签：problem ios else item ati class end case turn

题目描述

Youyouyouyou is very interested in math, one day, an idea came into his mind that how many ways he can patition n same things into no more than m groups? he think it too hard for him, so youyouyouyou ask wise cyt for help, but wise cyt don’t want to talk with youyouyouyou because it is too easy for him, now can you help youyouyouyou solve this problem?

输入

multi test cases, two integers n and m(1<=m<=n<=1000) , end with n = m = 0.

输出

output

#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
#define MOD 1000000007
typedef long long LL;
LL f[1001][1001] = {0},ans[1001][1001] = {0};
int main(){
    int n,m,i,j;
    for (i=1;i<=1000;i++){
        f[i][0] = 0;
        f[i][1] = 1;
        f[i][i] = 1;
        f[0][i] = 0;
    }
    for (i=1;i<=1000;i++){
        for (j=1;j<=i;j++){
            if (i == j) f[i][j] = 1;
            else 
                f[i][j] = (f[i-1][j-1] + f[i-j][j]) % MOD;
        }
    }
    for (i=1;i<=1000;i++){
        ans[i][1] = f[i][1];
        for (j=2;j<=i;j++){
            ans[i][j] = (ans[i][j-1] + f[i][j]) % MOD;
        }
    }
    while (scanf("%d %d",&n,&m) != EOF){
        if (n == 0 && m == 0) break;
//        LL res = 0;
//        for (i=1;i<=m;i++){
//            res = (res + f[n][i]) % MOD;
//        }
//        printf("%lld\n",res);
        printf("%lld\n",ans[n][m]);
    }
    return 0;
}