CodeForces - 287B-Pipeline（二分）

最新推荐文章于 2025-04-16 16:01:55 发布

转载最新推荐文章于 2025-04-16 16:01:55 发布 · 149 阅读

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原文链接：http://www.cnblogs.com/Staceyacm/p/10781882.html

本文探讨了在终极索伦城市中，通过使用不同类型的分流器来构建满足特定条件的水管网络的问题。主要关注如何利用最少数量的分流器，使城市中的每个房屋都能连接到供水系统，提供了一个算法解决方案，并通过实例展示了算法的应用。

Vova, the Ultimate Thule new shaman, wants to build a pipeline. As there are exactly n houses in Ultimate Thule, Vova wants the city to have exactly n pipes, each such pipe should be connected to the water supply. A pipe can be connected to the water supply if there's water flowing out of it. Initially Vova has only one pipe with flowing water. Besides, Vova has several splitters.

A splitter is a construction that consists of one input (it can be connected to a water pipe) and x output pipes. When a splitter is connected to a water pipe, water flows from each output pipe. You can assume that the output pipes are ordinary pipes. For example, you can connect water supply to such pipe if there's water flowing out from it. At most one splitter can be connected to any water pipe.

The figure shows a 4-output splitter

Vova has one splitter of each kind: with 2, 3, 4, ..., k outputs. Help Vova use the minimum number of splitters to build the required pipeline or otherwise state that it's impossible.

Vova needs the pipeline to have exactly n pipes with flowing out water. Note that some of those pipes can be the output pipes of the splitters.

Input

The first line contains two space-separated integers n and k (1 ≤ n ≤ 1018, 2 ≤ k ≤ 109).

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Output

Print a single integer — the minimum number of splitters needed to build the pipeline. If it is impossible to build a pipeline with the given splitters, print -1.

Examples

Input

4 3

Output

Input

5 5

Output

Input

8 4

Output

-1

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

long long n,k;
bool f(long long mid) {
	long long s;
	s=(k+1)*k/2-mid*(mid-1)/2-(k-mid);
	if(mid==k) s=mid;
	if(s>=n) return true;
	else return false;
}
int main() {

	scanf("%lld%lld",&n,&k);
	long long sum=0;
	sum=k*(k+1)/2-k+1;
	long long ans=k-1;
	long long l=2,r=k+1;
	long long mid=(l+r)/2;
	if(sum<n) {
		printf("-1\n");
	} else if(n==1) {
		printf("0");
	} else {
		while(mid-l>0) {
			//mid=(l+r)/2;
			if(f(mid)) {
				l=mid;
				mid=l+(r-l)/2;
			} else {
				mid=l+(mid-l)/2;
			}
		}
		long long ans=0;
		ans=k+1-mid;
		printf("%lld\n",ans);
	}
     return 0;
}

转载于:https://www.cnblogs.com/Staceyacm/p/10781882.html