lightoj1064 - Throwing Dice【dp打表】

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1064 - Throwing Dice

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Time Limit: 2 second(s)Memory Limit: 32 MB

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

Output for Sample Input

3 9

1 7

24 24

15 76

24 143

23 81

7 38

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

/* ***********************************************Author       : rycCreated Time : 2016-08-25 ThursdayFile Name    : E:\acm\LightOJ\106.cppLanguage     : c++Copyright 2016 ryc All Rights Reserved************************************************ */#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>using namespace std;typedef long long LL;typedef pair<int,int>pii;const int maxn=10010;LL dp[30][155];int n,m;LL gcd(LL a,LL b){    return b==0?a:gcd(b,a%b);}void Find(){    dp[0][0]=1ll;    for(int i=1;i<=25;++i){        for(int j=1;j<=150;++j){            for(int k=1;k<=6&&j-k>=0;++k){                dp[i][j]+=dp[i-1][j-k];            }        }    }}int main(){    Find();    int t,T=1;cin>>t;    while(t--){        scanf("%d%d",&n,&m);        LL ans=0,sum=1ll;        for(int i=n;i<m;++i){            ans+=dp[n][i];        }        for(int i=1;i<=n;++i){            sum*=6ll;        }ans=sum-ans;        printf("Case %d: ",T++);        if(ans==0){            printf("0\n");        }        else {            LL g=gcd(ans,sum);            if(sum/g==1){                printf("%lld\n",ans/g);            }            else {                printf("%lld/%lld\n",ans/g,sum/g);            }        }    }    return 0;}

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