UVA 11971 Polygon

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

思路：首先题目里面给的N是没有意义的...我们假设线段总长度是1.其次我们要知道假设多边形的K+1条边的长度分别为x1<=x2<=x3.....xk+1, 那么则有x1+x2+x3+...+xk>xk-1，即x1+x2+....+xk > 0.5。根据这个方程不好算，我们算它的反面，也就是说有且仅有一条边的长度不小于0.5。

于是：






推导过程就自己来吧....有规律的

代码：

#include <iostream>
#include <string.h>
#include <cstdio>
#include <cstring>
#include <cassert>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <cmath>
#include <algorithm>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<(int)(b);++i)
#define rrep(i,b,a) for(int i=(b);i>=(int)(a);--i)
#define clr(a,x) memset(a,x,sizeof(a))
#define ll long long
#define eps 1e-13

int main()
{
#ifdef ACM
freopen("in.txt","r",stdin);
#endif
int T; cin >> T;
rep(cas,1,T+1) {
int N, K; scanf("%d%d",&N,&K);
printf("Case #%d: ",cas);
if (K == 1) puts("0/1");
else {
ll fenzi = (K+1);
ll fenmu = 1LL << K;
fenzi = fenmu - fenzi;
ll g = __gcd(fenzi,fenmu);
fenzi /= g; fenmu /= g;
printf("%lld/%lld\n",fenzi,fenmu);
}
}
}