【LightOJ】1064 - 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

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

一直在想排列与组合的问题，比赛快结束想到了dp然而没时间了，挺亏的。

代码如下：

#include <cstdio>#include <stack>#include <cmath>#include <queue>#include <vector>#include <cstring>#include <algorithm>using namespace std;#define CLR(a,b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define LL long longLL GCD(LL a,LL b){return b == 0 ? a : GCD(b , a%b);}int main(){LL dp[26][155] = {0};for (int i = 6 ; i ; i--)dp[1][i] = 1;for (int i = 2 ; i <= 25 ; i++){for (int j = i * 6 ; j >= i ; j--){for (int k = j-1 ; k >= j - 6 && k > 0 ; k--)dp[i][j] += dp[i-1][k];}}int u;int n,x;int Case = 1;scanf ("%d",&u);while (u--){scanf ("%d %d",&n,&x);printf ("Case %d: ",Case++);if (x > n*6){puts("0");continue;}else if (x <= n){puts("1");continue;}LL sum1 = 0;for (int i = x ; i <= n*6 ; i++)sum1 += dp[n][i];LL sum2 = pow(6.0,n);LL g = GCD(sum1 , sum2);printf ("%lld/%lld\n",sum1/g,sum2/g);}return 0;}