LightOJ 1064 Throwing Dice

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

7

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

 

题意：给定n个骰子和一个x，要求出用这些骰子投出大于等于x的概率。要求最简。

分析：先预处理算出i个骰子得到的和为j的种数。然后用gcd函数约分。

dp[i+1][j+k]=sigma(dp[i][j]) (1<=k<=6)

#include <cstdio>using namespace std;__int64 dp[25][150];__int64 gcd(__int64 x, __int64 y){    return y ? gcd(y, x % y) : x;}int main(){    __int64 up, down, g;    int t, n, x;    for (int i = 1; i <= 25; i++){        for (int j = 1; j <= 150; j++){            if (i == 1 && j <= 6)                dp[i][j] = 1;            for (int k = 1; k <= 6; k++){                if (j >= k)                    dp[i][j] += dp[i - 1][j - k];            }        }    }    scanf("%d", &t);    for (int cas = 1; cas <= t; cas++){        scanf("%d%d", &n, &x);        up = 0, down = 0;        for (int i = n; i <= n * 6; i++){            down += dp[n][i];            if (x <= i)                up += dp[n][i];        }        if (up == down)            printf("Case %d: 1\n", cas);        else if (up == 0)            printf("Case %d: 0\n", cas);        else{            g = gcd(up, down);            printf("Case %d: %lld/%lld\n", cas, up / g, down / g);        }    }    return 0;}