UVA 10759 Dice Throwing(dp 概率)

Problem A
Dice Throwing
Input: standard input
Output: standard output
Time Limit: 1 second

 

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 0 1 11703055/78364164096 789532654692658645/789730223053602816 25/4738381338321616896 1/2 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。

思路：先用dp打表出用n个骰子掷出x的种数，然后就是用gcd约分。

代码：

#include <stdio.h>
#include <string.h>
const int N = 30;
const int X = 155;
long long n, x;
long long dp[N][X], zi, mu;

long long gcd(long long a, long long b) {
	if (!b) return a;
	return gcd(b, a % b); 
}
int main() {
	for (int i = 1; i <= 24; 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];
			}
		}
	while (~scanf("%lld%lld", &n, &x) && n || x) {
		mu = zi = 0;
		for (int i = n; i <= n * 6; i ++) {
			mu += dp[n][i];
			if (i >= x)
				zi += dp[n][i];
		}
		long long num = gcd(mu, zi);
		if (zi == mu)
			printf("1\n");
		else if (zi == 0)
			printf("0\n");
		else
			printf("%lld/%lld\n", zi / num, mu / num);
	}
	return 0;
}