Dice (II) （DP）唉，当时没做出来

Dice (II)

Time Limit: 3000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

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Description

You have N dices; each of them has K faces numbered from 1 to K. Now you can arrange the N dices in a line. If the summation of the top faces of the dices is S, you calculate the score as the multiplication of all the top faces.

Now you are given N, K, S; you have to calculate the summation of all the scores.

Input

Input starts with an integer T (≤ 25), denoting the number of test cases.

Each case contains three integers: N (1 ≤ N ≤ 1000) K (1 ≤ K ≤ 1000) S (0 ≤ S ≤ 15000).

Output

For each case print the case number and the result modulo 100000007.

Sample Input

5

1 6 3

2 9 8

500 6 1000

800 800 10000

2 100 10

Sample Output

Case 1: 3

Case 2: 84

Case 3: 74335590

Case 4: 33274428

Case 5: 165

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define mod 100000007
int dp[2][15001];
int main()
{
    long long t,r,n,k,s,j,kk,i;
    cin>>t;
    for(r=1; r<=t; r++)
    {
        memset(dp,0,sizeof(dp));
        cin>>n>>k>>s;
        for(j=1; j<=k; j++)
            dp[1][j]=j;
        long long size,sum,jia;
        for(i=2; i<=n; i++)
        {
            size=i*k>s?s:i*k;
            sum=0,jia=0;
            memset(dp[i&1],0,sizeof(dp[i&1]));
            for(j=i-1; j>=1&&i-j<=k; j--)
            {
                jia+=dp[(i-1)&1][j];
                sum+=dp[(i-1)&1][j]*(i-j);
            }
            for(j=i; j<=size; j++)
            {
                dp[i&1][j]=sum%mod;
                if(j-k>=1)
                {
                    jia-=dp[(i-1)&1][j-k];
                    sum-=dp[(i-1)&1][j-k]*k;
                }
                jia+=dp[(i-1)&1][j];
                sum+=jia;
            }
        }
        printf("Case %d: ",r);
        printf("%d
",dp[n&1][s]);
    }
}