CodeForces - 362B Petya and Staircases

最新推荐文章于 2019-02-22 22:38:03 发布

Lngxling

最新推荐文章于 2019-02-22 22:38:03 发布

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本文探讨了Petya如何在避开脏楼梯的情况下从第一级楼梯跳到第n级的问题。通过输入楼梯总数及脏楼梯的位置，算法判断是否存在一条只包含干净楼梯的路径。文章提供了一种解决方案，包括读取输入、排序脏楼梯位置并检查连续脏楼梯数量的过程。

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Little boy Petya loves stairs very much. But he is bored from simple going up and down them — he loves jumping over several stairs at a time. As he stands on some stair, he can either jump to the next one or jump over one or two stairs at a time. But some stairs are too dirty and Petya doesn't want to step on them.

Now Petya is on the first stair of the staircase, consisting of n stairs. He also knows the numbers of the dirty stairs of this staircase. Help Petya find out if he can jump through the entire staircase and reach the last stair number n without touching a dirty stair once.

One has to note that anyway Petya should step on the first and last stairs, so if the first or the last stair is dirty, then Petya cannot choose a path with clean steps only.

Input

The first line contains two integers n and m (1 ≤ n ≤ 109, 0 ≤ m ≤ 3000) — the number of stairs in the staircase and the number of dirty stairs, correspondingly. The second line contains m different space-separated integers d1, d2, ..., dm (1 ≤ di ≤ n) — the numbers of the dirty stairs (in an arbitrary order).

Output

Print "YES" if Petya can reach stair number n, stepping only on the clean stairs. Otherwise print "NO".

Examples

input
10 5
2 4 8 3 6

output
NO

input
10 5
2 4 5 7 9

output
YES

数出最多连续的凳子数，注意脏的凳子数可能为0

#include <iostream>
#include <cstdio>
#include <algorithm>
#define max_ 3010
using namespace std;
int n,m;
int num[max_];
int main()
{
	cin>>n>>m;
	for(int i=1;i<=m;i++)
	{
		cin>>num[i];
	}
	if(m==0)
	{
		printf("YES\n");
		return 0;
	}
	sort(num+1,num+1+m);
	if(num[1]==1||num[m]==n)
	{
		printf("NO\n");
		return 0;
	}
	int i,cnt=1;
	for(i=2;i<=m;i++)
	{
		if(num[i]-num[i-1]==1)
		{
			cnt++;
			if(cnt>2)
			break;
		}
		else
		{
			cnt=1;
		}
	}
	if(i==m+1)
	printf("YES\n");
	else
	printf("NO\n");
}