lightoj-1193 - Dice (II)(dp+前缀和)

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

 

解题思路: 跟(I)一样分析可得dp[i][j] = dp[i-1][j-1]+dp[i-1][j-2]*2 + ……+dp[i-1][j-k]*k;

则dp[i][j-1] = dp[i-1][j-2]+dp[i-1][j-3]*2 + ……+dp[i-1][j-k-1]*k;

比较可得dp[i][j]=dp[i][j-1] - dp[i-1][j-k-1]*k + sum[i-1][j-1]-sum[i-1][j-k-1];

sum[i][j]为dp[i][j] 的第i层的前j项和。

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

const int mod = 100000007;
const int N = 1010,S = 15010;
long long dp[2][S],sum[2][S];

int main(){
    
    int T,n,k,s;
    scanf("%d",&T);
    for(int t=1;t<=T;t++){
        
        memset(dp,0,sizeof(dp));
        memset(sum,0,sizeof(sum));
        scanf("%d%d%d",&n,&k,&s);
        for(int i=1;i<=k;i++) dp[1][i] = i,sum[1][i] = sum[1][i-1]+i;
        for(int i=k+1;i<=s;i++) sum[1][i] = sum[1][i-1];
        for(int i=2;i<=n;i++){
            for(int j=1;j<=s;j++){
                dp[i%2][j] = (dp[i%2][j-1] + sum[(i+1)%2][j-1] - sum[(i+1)%2][max(0,j-k-1)]-(dp[(i+1)%2][max(0,j-k-1)]*k)%mod+mod)%mod;
                sum[i%2][j] = sum[i%2][j-1] + dp[i%2][j];
            }
        }
        printf("Case %d: %lld\n",t,dp[n%2][s]);
    }
    
    return 0;
}