Light oj 1145 - Dice (I)(dp 递推)

1145 - Dice (I)

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Time Limit: 2 second(s)Memory Limit: 32 MB

You have N dices; each of them has K faces numbered from 1 to K. Now you have arranged the N dices in a line. You can rotate/flip any dice if you want. How many ways you can set the top faces such that the summation of all the top faces equals S?

Now you are given N, K, S; you have to calculate the total number of ways.

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) and 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: 1

Case 2: 7

Case 3: 57286574

Case 4: 72413502

Case 5: 9

 

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PROBLEM SETTER: JANE ALAM JAN

/*题意:  n个色子，每个有k面，求摆放成  总和为  s 的情况数  mod  100000007借鉴别人思路 dp[i][v]  代表前  i 个色子 组合成  v 的情况，那么 dp[i][v]=dp[i][v-1]+dp[i-1][v]- ( v-k-1>=0: dp[i-1][v-k-1]:0)   后面是说 前面的状态可以通过 这一个塞子的点数得到*/#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define bug printf("hihi\n")#define eps 1e-12typedef long long ll;using namespace std;#define mod 100000007#define INF 0x3f3f3f3f#define N  15101ll dp[2][N];int n,k,s;int ca=0;int get(int st,int v){      int te=dp[st][v-1];      if(v-k-1>=0) te-=dp[st][v-k-1];      return te;}void DP(){    int i,j;    memset(dp,0,sizeof(dp));    int cur=0;    dp[0][1]=1;    for(int i=2;i<=s;i++)    {        dp[0][i]=dp[0][i-1];        if(i<=k) dp[0][i]+=1;    }    for(i=1;i<n;i++)    {        int to=cur^1;        memset(dp[to],0,sizeof(dp[to]));        for(int v=1;v<=s;v++)        {            dp[to][v]+=dp[to][v-1];            dp[to][v]%=mod;            dp[to][v]+=get(cur,v);            dp[to][v]%=mod;        }        cur^=1;    }    printf("Case %d: %lld\n",++ca,(dp[cur][s]-dp[cur][s-1]+mod)%mod);}int main(){    int i,j,t;    ca=0;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%d",&n,&k,&s);        if(s==0)        {            printf("Case %d: 0\n",++ca);            continue;        }        DP();    }   return 0;}