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

題解：dp[ i ][ j ]表示 i 個骰子擲出和為 j 的方案數。

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

 1 #define INF 1e8
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<iostream>
 5 #include<algorithm>
 6 using namespace std;
 7 typedef long long
 
 ll;
 8 
 9 int n,x;
10 ll sum;
11 ll dp[25][150];
12 
13 ll gcd(ll a,ll b){
14     return b==0?a:gcd(b,a%b);
15 }
16 
17 void solve(){
18     sum=1;
19     memset(dp,0,sizeof(dp));
20     dp[0][0]=1;
21     for(int i=1;i<=n;i++){
22         sum=sum*6;
23         for(int j=1;j<=i*6;j++)
24             for(int k=1;k<=6;k++)
25                 if(j>=k) dp[i][j]=dp[i][j]+dp[i-1][j-k];          //因為dp[0][0]=1,所以j一定要大於等於k，否則不能更新dp[i][j]的值
26     }
27 }
28 
29 int main()
30 {   int kase;
31     cin>>kase;
32     for(int t=1;t<=kase;t++){
33         cin>>n>>x;
34         solve();
35         ll ans=0;
36         for(int i=x;i<=n*6;i++) ans+=dp[n][i];
37         ll temp=gcd(sum,ans);
38         if(ans==0) printf("Case %d: 0\n",t);
39         else if(ans==sum)     printf("Case %d: 1\n",t);
40         else printf("Case %d: %lld/%lld\n",t,ans/temp,sum/temp);
41     }
42 }

Throwing Dice LightOJ - 1064