Election

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本文介绍了一个选举预测的问题，通过数学概率来提前判断候选人的胜算。文章提供了完整的代码实现，使用了高中水平的排列组合知识来解决该问题。

F(1967): Election

Time Limit: 1 Sec Memory Limit: 128 Mb Submitted: 495 Solved: 44

Description

After all the fundraising, campaigning and debating, the election day has finally arrived. Only two candidates remain on the ballot and you work as an aide to one of them.

Reports from the voting stations are starting to trickle in and you hope that you can soon declare a victory.

There are N voters and everyone votes for one of the two candidates (there are no spoiled ballots). In order to win, a candidate needs more than half of the votes. A certain number M ≤ N of ballots have been counted, and there are Vi votes for candidate i (V1+V2 = M), where V1 is the number of votes your candidate received.

Due to the historical data and results of highly scientific polls, you know that each of the remaining votes has a 50% chance to go to your candidate. That makes you think that you could announce the win before all the votes are counted. So, if the probability of winning strictly exceeds a certain threshold W, the victory is yours! We just hope you are sure of this, we don’t want any scandals...

Input

The first line of input contains a single positive integer T ≤ 100 indicating the number of test cases. Next T lines each contain four integers: N, V1, V2 and W as described above.

The input limits are as follows:

1 ≤ N ≤ 50
50 ≤ W < 100
V1, V2 ≥ 0
V1 + V2 ≤ N

Output

For each test case print a single line containing the appropriate action:

If the probability that your candidate will win is strictly greater than W%, print GET A CRATE OF CHAMPAGNE FROM THE BASEMENT!
If your candidate has no chance of winning, print RECOUNT!
Otherwise, print PATIENCE, EVERYONE!

Sample Input

Sample Output

RECOUNT!
PATIENCE, EVERYONE!
PATIENCE, EVERYONE!
GET A CRATE OF CHAMPAGNE FROM THE BASEMENT!

题意：如果您的候选人获胜的概率严格大于W％，输出 GET A CRATE OF CHAMPAGNE FROM THE BASEMENT!

如果您的候选人没有获胜的机会，输出 RECOUNT!

否则，输出  PATIENCE, EVERYONE!

直接贴代码：/*
F题：
高中的排列组合知识就可以解决，从n个物品中选择m个物品,每个物品被选择的概率都是0.5；
C(m,n)*（0.5）^m(选中)*(0.5)^(n-m)（未选中）=C(m,n)*(0.5)^n;
根据这个公式，再判断v1得到几票可以赢。
*/
#include<cstdio>
#include<cmath>
using namespace std;
typedef long long ll;
ll C(int n, int m)
{
	ll ans = 1;
	ll ans2 = 1;
	int i, j;
	for (i = n, j = m; i>n - m; i--, j--)
	{
		ans *= i;
		if (ans%j == 0)
			ans /= j;
		else
			ans2 *= j;
	}
	return ans / ans2;
}
int main()
{
	int n, v1, v2, t;
	double w;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d%d%d%lf", &n, &v1, &v2, &w);
		double sum = 0;
		w /= 100.0;
		for (int i = n / 2 + 1 - v1; i <= n - v1 - v2; i++)
		{
			sum += 1.0*C(n - v1 - v2, i)*pow(0.5, n - v1 - v2);//重点知识 
		}
		if (v2 == n / 2 && n % 2 == 0)
			printf("RECOUNT!\n");//特判 
		else
		{
			if (v1 >= (n / 2 + 1) || sum>w)
				printf("GET A CRATE OF CHAMPAGNE FROM THE BASEMENT!\n");
			else if (v2 >= (n / 2 + 1))
				printf("RECOUNT!\n");
			else
				printf("PATIENCE, EVERYONE!\n");
		}
	}
	return 0;
}