UVa 10759 Dice Throwing (概率DP)-优快云博客

本文链接：https://blog.youkuaiyun.com/synapse7/article/details/12241351

本文介绍了一种计算掷多个常规立方体骰子后得到的点数总和至少为某一数值的概率的方法。通过动态规划的方式逐步计算所有可能的结果，并最终以最简分数形式给出答案。

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10759 - Dice Throwing

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1700

n common cubic dice are thrown. What is the probability that the sum of all thrown dice is at least x?

Input

The input file contains several test cases. Each test case consists two integers n (1<=n<=24) and x(0<=x<150). The meanings of n and x are given in the problem statement. Input is terminated by a case where n=0 and x=0. This case should not be processed.

Output

For each line of input produce one line of output giving the requested probability as a proper fraction in lowest terms in the format shown in the sample output. All numbers appearing in output are representable in unsigned 64-bit integers. The last line of input contains two zeros and it should not be processed.

Sample Input Output for Sample Input

3 9

1 7

24 24

15 76

24 56

24 143

23 81

7 38

0 0

20/27

11703055/78364164096

789532654692658645/789730223053602816

25/4738381338321616896

1/2

55/46656

从下往上慢慢加~

f[i + 1][j + k] += f[i][j];

完整代码：

/*0.015s*/

#include<cstdio>
typedef unsigned long long ull;

ull f[25][150];

ull gcd(ull a, ull b)
{
	return b ? gcd(b, a % b) : a;
}

int main()
{
	int i, j, k, n, x;
	ull ans, b, g;
	for (i = 1; i <= 6; ++i)
		f[1][i] = 1;
	for (i = 1; i < 25; ++i)
		for (j = i; j <= 6 * i; ++j)
			for (k = 1; k <= 6; ++k)
				f[i + 1][j + k] += f[i][j];
	while (scanf("%d%d", &n, &x), n)
	{
		if (x > 6 * n)
		{
			puts("0");
			continue;
		}
		else if (x <= n)
		{
			puts("1");
			continue;
		}
		ans = 0, b = 1;
		for (i = x; i <= 6 * n; ++i) ans += f[n][i];
		for (i = 0; i < n; ++i) b *= 6;
		g = gcd(ans, b);
		printf("%llu/%llu\n", ans / g, b / g);
	}
}