HDOJ 题目4284 A Famous Stone Collector（组合数学）_hdoj的经典组合数学题-优快云博客

本文介绍了一道关于组合数学的问题，即如何计算不同颜色石头的所有可能排列方式。问题要求使用编程方法解决，输入为每种颜色石头的数量，输出为所有可能的不同排列数量。

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A Famous Stone Collector

Time Limit: 30000/15000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1122 Accepted Submission(s): 421

Problem Description

Mr. B loves to play with colorful stones. There are n colors of stones in his collection. Two stones with the same color are indistinguishable. Mr. B would like to
select some stones and arrange them in line to form a beautiful pattern. After several arrangements he finds it very hard for him to enumerate all the patterns. So he asks you to write a program to count the number of different possible patterns. Two patterns are considered different, if and only if they have different number of stones or have different colors on at least one position.

Input

Each test case starts with a line containing an integer n indicating the kinds of stones Mr. B have. Following this is a line containing n integers - the number of
available stones of each color respectively. All the input numbers will be nonnegative and no more than 100.

Output

For each test case, display a single line containing the case number and the number of different patterns Mr. B can make with these stones, modulo 1,000,000,007,
which is a prime number.

Sample Input

Sample Output

  
   Case 1: 15
Case 2: 8


   
    
     Hint
    
In the first case, suppose the colors of the stones Mr. B has are B, G and M, the different patterns Mr. B can form are: B; G; M; BG; BM; GM; GB; MB; MG; 
BGM; BMG; GBM; GMB; MBG; MGB.

Source

Fudan Local Programming Contest 2012

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题目大意：n用颜色的球，给出每种球的个数，问总共有多少种排法，石头可以不用上

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define LL long long
#define mod 1000000007
using namespace std;
LL a[110];
LL c[11010][110];
int dp[110][11010];
void fun()
{
    int i,j;
    for(i=0;i<10010;i++)
        c[i][0]=1;
    for(i=1;i<10010;i++)
        for(j=1;j<=i&&j<110;j++)
        c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;
}
int main()
{
    LL n;
    int cas=0;
    fun();
    while(scanf("%lld",&n)!=EOF)
    {
        LL i,j,k;
        LL sum=0;
        int m;
       // scanf("%d",&m);
       // printf("++++%lld\n",c[n][m]);
        for(i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            sum+=a[i];
        }
        dp[0][0]=1;
        for(i=1;i<=n;i++)
        {
            for(j=0;j<=sum;j++)
            {
                dp[i][j]=dp[i-1][j];
                for(k=1;k<=a[i];k++)
                {
                    if(j-k<0)
                        break;
                    dp[i][j]=(dp[i][j]+dp[i-1][j-k]*c[j][k]%mod)%mod;
                }
            }
        }
        LL ans=0;
        for(i=1;i<=sum;i++)
        {
            ans=(ans+dp[n][i])%mod;
        }
        printf("Case %d: %lld\n",++cas,ans);
    }
}