Throwing Dice LightOJ - 1064

n common cubic dice are thrown. What is the probability that the sum of all thrown dice is at least x?

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each test case contains two integers n (1 ≤ n < 25) and x (0 ≤ x < 150). The meanings of n and x are given in the problem statement.

Output

For each case, output the case number and the probability in 'p/q' form where p and q are relatively prime. If q equals 1 then print p only.

Sample Input

7

3 9

1 7

24 24

15 76

24 143

23 81

7 38

Sample Output

Case 1: 20/27

Case 2: 0

Case 3: 1

Case 4: 11703055/78364164096

Case 5: 25/4738381338321616896

Case 6: 1/2

Case 7: 55/4665

题解：dp[ i ][ j ]表示 i 个骰子掷出和为 j 的方案数。

所以，dp[ i ][ j ]=dp[ i ][ j ]+dp[ i - 1 ][ j - k ](1<=k<j)

#define INF 1e8
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

int n,x;
ll sum;
ll dp[25][150];

ll gcd(ll a,ll b){
    return b==0?a:gcd(b,a%b);
}

void solve(){
    sum=1;
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    for(int i=1;i<=n;i++){
        sum=sum*6;
        for(int j=1;j<=i*6;j++)
            for(int k=1;k<=6;k++)
                if(j>=k) dp[i][j]=dp[i][j]+dp[i-1][j-k];          //因为dp[0][0]=1,所以j一定要大于等于k，否则不能更新dp[i][j]的值
    }
}

int main()
{   int kase;
    cin>>kase;
    for(int t=1;t<=kase;t++){
        cin>>n>>x;
        solve();
        ll ans=0;
        for(int i=x;i<=n*6;i++) ans+=dp[n][i];
        ll temp=gcd(sum,ans);
        if(ans==0) printf("Case %d: 0\n",t);
        else if(ans==sum)     printf("Case %d: 1\n",t);
        else printf("Case %d: %lld/%lld\n",t,ans/temp,sum/temp);
    }
}