Throwing Dice LightOJ-1064

Tag: code greater than Mon void lightoj type Test bic names

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

Input

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

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

Output

For each case, output the case number and the probability in ' p/q ' form where p and Q is Relat Ively 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

Solution: Dp[I [j] represents the number of scenarios where I dice are thrown and are J.

So, dp[I [J]=dp[i] [j]+dp[I-1] [j-k] (1<=K<J)

1 #defineINF 1e82#include <cstdio>3#include <cstring>4#include <iostream>5#include <algorithm>6 using namespacestd;7typedefLong Longll;8 9 intn,x;Ten ll sum; Onell dp[ -][ Max]; A  - ll gcd (ll A,ll b) { -     returnb==0? A:GCD (b,a%b); the } -  - voidsolve () { -sum=1; +Memset (DP,0,sizeof(DP)); -dp[0][0]=1; +      for(intI=1; i<=n;i++){ Asum=sum*6; at          for(intj=1; j<=i*6; j + +) -              for(intk=1; k<=6; k++) -                 if(j>=k) dp[i][j]=dp[i][j]+dp[i-1][j-K]; Because Dp[0][0]=1, J must be greater than or equal to K, otherwise the value of Dp[i][j] cannot be updated -     } - } -  in intMain () -{intKase; toCin>>Kase; +      for(intt=1; t<=kase;t++){ -Cin>>n>>x; the solve (); *ll ans=0; $          for(inti=x;i<=n*6; i++) ans+=Dp[n][i];Panax Notoginsengll temp=gcd (Sum,ans); -         if(ans==0) printf ("Case %d:0\n", T); the         Else if(ans==sum) printf ("Case %d:1\n", T); +         Elseprintf"Case %d:%lld/%lld\n", t,ans/temp,sum/temp); A     } the}

Throwing Dice LightOJ-1064