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UVA 10759 Dice Throwing(dp 概率)

2013-11-07 22:15 381 查看

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 [b]x?[/b]

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

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
[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
[i];
if (i >= x)
zi += dp
[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;
}

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