Dice (I)

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

Source

Jane Alam Jan

题意：给你n个骰子，每个骰子有1-k个分数，问你多少种方式的和是S

题解：  首先dp[i][j]表示前i个骰子，和为S的方案数

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

对于i完全由i-1的状态得到，我们这可以用   sum[j]表示 dp[i-1][1到j]的一个前缀和,

用i表示当前要计算的now状态，last表示i-1的状态 由此我们可以 得到dp[now][j]=sum[j-1]-sum[j-k-1];

最终答案就是dp[now][s];

///1085422276
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<bitset>
#include<set>
#include<vector>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define memfy(a)  memset(a,-1,sizeof(a));
#define TS printf("111111\n");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define mod 100000007
#define inf 100000000
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<‘0‘||ch>‘9‘)
    {
        if(ch==‘-‘)f=-1;
        ch=getchar();
    }
    while(ch>=‘0‘&&ch<=‘9‘)
    {
        x=x*10+ch-‘0‘;
        ch=getchar();
    }
    return x*f;
}
//****************************************

#define maxn 15000+5
ll dp[2][maxn],sum[maxn];
int n,k,s;
int main()
{

    int T=read();
    int oo=1;
    while(T--)
    {
        scanf("%d%d%d",&n,&k,&s);
          int last=0,now=1;
          mem(dp);
          dp[now][0]=1;
        for(int i=1;i<=n;i++)
        {
            swap(last,now);
            mem(dp[now]);
            sum[0]=dp[last][0];
           for(int j=1;j<=s;j++)
            {
                sum[j]=(sum[j-1]+dp[last][j])%mod;
            }
            for(int j=1;j<=s;j++)
            {
                if(j<=k) dp[now][j]=(sum[j-1])%mod;
                    else
                dp[now][j]=(sum[j-1]-sum[j-k-1]+mod)%mod;
            }
        }
        printf("Case %d: ",oo++);
        cout<<dp[now][s]<<endl;;
    }
    return 0;
}

代码