HDOJ 4248 A Famous Stone Collector DP

DP: dp[i][j]前i堆排出j长度的序列有多少种排法，

dp[i][j]=dp[i-1][j] (不用第i堆), 

dp[i][j]+=dp[i-1][j-k]*C[j][k] (用第i堆的k个石头)

A Famous Stone Collector

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 845    Accepted Submission(s): 322

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

   
   
   
   

    
    
    
    3
1 1 1
2
1 2
   
   
   
   

 

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

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long int LL;

const LL MOD = 1000000007LL;

LL C[11000][210];
void getC()
{
    for(int i=0;i<11000;i++) C[i][0]=C[i][i]=1;
    for(int i=2;i<11000;i++)
        for(int j=1;j<i&&j<=200;j++)
            C[i][j]=(C[i-1][j]+C[i-1][j-1])%MOD;
}

LL dp[110][11000];
int n,a[110];

int main()
{
    getC();
    int cas=1;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        memset(dp,0,sizeof(dp));
        int len=0;
        dp[0][0]=1;
        for(int i=1;i<=n;i++)
        {
            len+=a[i];
            for(int j=0;j<=len;j++)
            {
                dp[i][j]=(dp[i][j]+dp[i-1][j])%MOD;
                for(int k=1;k<=a[i]&&j-k>=0;k++)
                {
                    dp[i][j]=(dp[i][j]+dp[i-1][j-k]*C[j][k])%MOD;
                }
            }
        }
        LL ans=0;
        for(int i=1;i<=len;i++)
            ans=(ans+dp[n][i])%MOD;
        printf("Case %d: %lld\n",cas++,ans%MOD);
    }
    return 0;
}