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 Fis 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);
        }
    }
}