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[30][155];

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

void Init()
{
    for(int i=1;i<=6;i++)//一个骰子
        dp[1][i]=1;
    for(int i=2;i<25;i++)
    {
        for (int j=i;j<=6*i;j++)//i个骰子所有的点数
        {
            for (int k=1;k<=6;k++)
            {
                if (j-k>=0)
                dp[i][j]+=dp[i-1][j-k];
            }
        }
    }
}

int main()
{
    Init();
    int T;
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++)
    {
        scanf("%d%d",&n,&x);
        if(x>n*6)
        {
            printf("Case %d: 0
",cas);
            continue;
        }
        if(x<=n)
        {
            printf("Case %d: 1
",cas);
            continue;
        }
        LL up=0,down=1,g;
        for(int i=x;i<=n*6;i++)
            up+=dp[n][i];
        for(int j=0;j<n;j++) down*=6;
        g=gcd(up,down);
        printf("Case %d: %lld/%lld
",cas,up/g,down/g);
    }
    return 0;
}