多边形的组成条件是最长边不能占边长总和的一半,将木棒想象成圆多砍一刀,然后是简单概率.

Polygon Time Limit: 1000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status Description U   — Polygon Time Limit: 1 sec Memory Limit: 32 MB John has been given a segment of lenght N, however he needs a polygon. In order to create a polygon he has cut given segment K times at random positions (uniformly distributed cuts). Now he has K+1 much shorter segments. What is the probability that he can assemble a polygon using all new segments? INPUT The number of tests T (T ≤ 1000) is given on the first line. T lines follow, each of them contains two integers N K (1 ≤ N ≤ 106; 1 ≤ K ≤ 50) described above. OUTPUT For each test case output a single line "Case #T: F". Where T is the test case number (starting from 1) and F is the result as simple fraction in form of N/D. Please refer to the sample output for clarity. SAMPLE INPUT 2 1 1 2 2 SAMPLE OUTPUT Case #1: 0/1 Case #2: 1/4 Problem by: Aleksej Viktorchik; Leonid Sislo Huge Easy Contest #2 Source Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 10. Maths :: Examples Root :: Prominent Problemsetters :: Leonid Shishlo Root :: Prominent Problemsetters :: Aleksej Viktorchik Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 2. Mathematics :: Probability :: Exercises: Intermediate Submit Status

/* ***********************************************
Author        :CKboss
Created Time  :2015年01月30日 星期五 21时37分11秒
File Name     :UVA11971.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;
int n;

LL gcd(LL a,LL b)
{
	if(b!=0) return gcd(b,a%b);
	return a;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	int T_T,cas=1;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%*d%d",&n);
		LL fenzhi = (1LL<<n)-n-1;
		LL fenmu = (1LL<<n);
		LL g = gcd(fenzhi,fenmu);
		printf("Case #%d: %lld/%lld\n",cas++,fenzhi/g,fenmu/g);
	}
    
    return 0;
}

Uva 11971 Polygon 想法