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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

3
1 1 1
2
1 2

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.
 

#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;
}

HDOJ 4248 A Famous Stone Collector DP

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原文地址：http://blog.csdn.net/ck_boss/article/details/40209735