XidianOJ 1183 Water Problem: Items divided

Posted 2020-08-22 TOTOTOTOTZZZZZ

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;
} 

 

an answer modulo 1e9 + 7 per line

--正文

n个相同的球放入m个相同的盒子，允许有空盒

  F(n,m) = F(n-1,m-1) + F(n-m,m)

注意边界的处理