hdu 4248 A Famous Stone Collector(组合数学&DP)

A Famous Stone Collector

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 765    Accepted Submission(s): 286

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

 

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题意：

给定n种颜色的石头，每种颜色有num[i]颗,同种颜色的石头不区分。问能构成多少种不同的石头序列（不同的序列是指：1.石头数不同；2.石头数相同，至少一个位置的石头颜色不同）

思路：
首先抽象出状态 dp[ i ][ j ]表示前i种石头构成的长度为j的序列的个数。

那么当处理i种石头时。可以一个都不要。dp[i][j]=dp[i-1][j]。

也可以要k个。那么相当于将这k个个石头放在j个位置上。而将k个石头放在j个位置上的方法数有C[j][k]种。

而以前的j-k个还是按照原排列。所以有dp[i-1][j-k]*C[j][k]种方法。所以。

dp[i][j]=sum(dp[i-1][j-k]*C[j][k])。0<=k<=num[i]。

详细见代码：

#include<algorithm>
#include<iostream>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=100010;
const int mod=1000000007;
int C[15000][150],num[150];
__int64 dp[150][15000];//注意数组大小
int main()
{
    int i,j,k,n,cas=1,sum,ans;
    for(i=0;i<=10010;i++)//注意组合数的范围
    {
        C[i][0]=1;
        for(j=1;j<=i&&j<=105;j++)
            C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;
    }
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
            scanf("%d",num+i);
        sum=0;
        memset(dp,0,sizeof dp);
        dp[0][0]=1;
        for(i=1;i<=n;i++)
        {
            sum+=num[i];
            for(k=0;k<=num[i];k++)//由于利用上层结果。j,k嵌套顺序无关。
                for(j=k;j<=sum;j++)
                    dp[i][j]=(dp[i][j]+dp[i-1][j-k]*C[j][k])%mod;
        }
        ans=0;
        for(i=1;i<=sum;i++)
            ans=(ans+dp[n][i])%mod;
        printf("Case %d: %d\n",cas++,ans);
    }
    return 0;
}