Throwing Dice(概率dp)

C - Throwing Dice

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

LightOJ 1064 uDebug

Description

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: 55/46656

//比赛没看懂题，这是一个简单概率dp，第一行是案例数 T ，然后 n ， m 是n个骰子掷出至少为 m 的概率。

dp[i][j]代表 i 个骰子，掷出 j 种数

dp[i][j]=SUM(dp[i-1][j-k]) (1<=k<=6)

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 using namespace std;
 #define LL long long

 int n,x;
 LL dp[][];

 LL gcd(LL x,LL y)
 {
     return y?gcd(y,x%y):x;
 }

 void Init()
 {
     for(int i=;i<=;i++)//一个骰子
         dp[][i]=;
     for(int i=;i<;i++)
     {
         for (int j=i;j<=*i;j++)//i个骰子所有的点数
         {
             for (int k=;k<=;k++)
             {
                 if (j-k>=)
                 dp[i][j]+=dp[i-][j-k];
             }
         }
     }
 }

 int main()
 {
     Init();
     int T;
     scanf("%d",&T);
     for(int cas=;cas<=T;cas++)
     {
         scanf("%d%d",&n,&x);
         if(x>n*)
         {
             printf("Case %d: 0\n",cas);
             continue;
         }
         if(x<=n)
         {
             printf("Case %d: 1\n",cas);
             continue;
         }
         LL up=,down=,g;
         for(int i=x;i<=n*;i++)
             up+=dp[n][i];
         for(int j=;j<n;j++) down*=;
         g=gcd(up,down);
         printf("Case %d: %lld/%lld\n",cas,up/g,down/g);
     }
     return ;
 }