【LightOJ】1064 - Throwing Dice（dp打表）

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本文探讨了给定数量的普通立方骰子投掷后，总和至少为某个数值的概率计算问题。通过动态规划方法预先计算所有可能的结果，并提供了一个有效的解决方案。

1064 - Throwing Dice

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

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 long
LL 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;
}