1193 - Dice (II)

1193 - Dice (II)

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Time Limit: 3 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
5 1 6 3 2 9 8 500 6 1000 800 800 10000 2 100 10	Case 1: 3 Case 2: 84 Case 3: 74335590 Case 4: 33274428 Case 5: 165

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PROBLEM SETTER: JANE ALAM JAN

思路:和1145差不多。

dp[i][j]表示前i次点数之和为j的数所有对数的各个的乘积之和。

那么dp[i][j]=dp[i-1][j-1]+dp[i-1][j-2]*2+....dp[i-1][j-k]*k;

这样的复杂度是n*n*n;这样是过不了的；

那么dp[i][j+1]=dp[i-1][j]+dp[i-1][j-1]*2+dp[i-1][j-2]*3+....dp[i-1][j-k+1]*k;

那么dp[i][j+1]=dp[i][j-1]+dp[i-1][j-1]+sum[j-2]-sum[max(0,j-1-m)]-dp[i-1][max(0,j-m-1)]*m;

那么我们每次维护一个dp[i][j]的前缀和就可以了。

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<math.h>
 6 #include<stdlib.h>
 7 #include<queue>
 8 using namespace std;
 9 typedef long long LL;
10 LL dp[2][15005];
11 const int mod=100000007;
12 LL sum[15005];
13 int main(void)
14 {
15     int i,j,k;
16     scanf("%d",&k);
17     int s;
18     int n,m,c;
19     for(s=1; s<=k; s++)
20     {
21         scanf("%d %d %d",&n,&m,&c);
22         memset(dp,0,sizeof(dp));
23         memset(sum,0,sizeof(sum));
24         for(i=1; i<=m; i++)
25         {
26             dp[1][i]=i;
27             dp[1][i]%=mod;
28             sum[i]=(sum[i-1]+dp[1][i])%mod;
29         }
30         for(i=m+1;i<=c;i++)
31             sum[i]=sum[i-1];
32         for(i=2; i<=n; i++)
33         {
34             int uu=(i)%2;
35             int kk=(i+1)%2;
36             for(j=0; j<i; j++)
37             {
38                 dp[uu][j]=0;
39             }
40             for(j=i;j<=c; j++)
41             {
42       dp[uu][j]=((dp[kk][j-1]+dp[uu][j-1])%mod+(sum[j-2]-sum[max(0,j-1-m)])%mod-dp[kk][max(0,j-m-1)]*m%mod)%mod;
43                 dp[uu][j]%=mod;
44                 dp[uu][j]+=mod;
45                 dp[uu][j]%=mod;
46             }
47             for(j=1; j<=c; j++)
48             {
49                 sum[j]=sum[j-1]+dp[uu][j];
50                 sum[j]%=mod;
51             }
52         }
53         printf("Case %d: ",s);
54         printf("%lld\n",(dp[n%2][c]%mod+mod)%mod);
55     }
56     return 0;
57 }