lightoj 1064 - Throwing Dice

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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

这题的大意是给你n个骰子，问你掷出不小于x的概率是多少。

因为所有点数的个数是6^n，所以只用求出不小于x的种数有多少个就可以了。

#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define inf 1e9#define LL long longusing namespace std;LL dp[30][160];LL gcd(LL a,LL b){    if(b == 0)        return a;    else        return gcd(b,a%b);}int main(void){    int T,n,x,i,j,k;    scanf("%d",&T);    int cas = 1;    while(T--)    {        scanf("%d%d",&n,&x);        memset(dp,0,sizeof(dp));        dp[0][0] = 1;        LL sum = 1;        for(i=1;i<=n;i++)        {            sum *= 6;            for(j=1;j<=i*6;j++)            {                for(k=1;k<=6;k++)                {                    if(j < k)                        break;                    dp[i][j] += dp[i-1][j-k];                }            }        }        LL ans = 0;        for(i=x;i<=n*6;i++)            ans += dp[n][i];        printf("Case %d: ",cas++);        if(ans == 0)        {            printf("0\n");            continue;        }        if(ans == sum)        {            printf("1\n");            continue;        }        LL t = gcd(ans,sum);        printf("%lld/%lld\n",ans/t,sum/t);    }}

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