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[][];
 int main()
 {
     long long t,r,n,k,s,j,kk,i;
     cin>>t;
     for(r=; r<=t; r++)
     {
         memset(dp,,sizeof(dp));
         cin>>n>>k>>s;
         for(j=; j<=k; j++)
             dp[][j]=j;
         long long size,sum,jia;
         for(i=; i<=n; i++)
         {
             size=i*k>s?s:i*k;
             sum=,jia=;
             memset(dp[i&],,sizeof(dp[i&]));
             for(j=i-; j>=&&i-j<=k; j--)
             {
                 jia+=dp[(i-)&][j];
                 sum+=dp[(i-)&][j]*(i-j);
             }
             for(j=i; j<=size; j++)
             {
                 dp[i&][j]=sum%mod;
                 if(j-k>=)
                 {
                     jia-=dp[(i-)&][j-k];
                     sum-=dp[(i-)&][j-k]*k;
                 }
                 jia+=dp[(i-)&][j];
                 sum+=jia;
             }
         }
         printf("Case %d: ",r);
         printf("%d\n",dp[n&][s]);
     }
 }