CodeForces 362B Petya and Staircases

小咪咪 2022-08-22 15:29 5阅读 0赞

# CodeForces 362B Petya and Staircases  #

**time limit per test 1 second**

**memory limit per test 256 megabytes**

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 *d*1, *d*2, ..., *d**m* (1 ≤ *d**i* ≤ *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 ##

## Input ##

10 5
    2 4 5 7 9

## Output ##

YES

链接在这：[题目链接][Link 1]

排序，遍历寻找连续的dirty是否超过2个

代码：

#include<stdio.h>
    #include<algorithm>
    using namespace std;
    
    int a[3005];
    
    int main()
    {
        int n,m;
        while(~scanf("%d %d",&n,&m))
        {
            for(int i=0; i<m; i++)
                scanf("%d",&a[i]);
            if(m>=n)
            {
                printf("NO\n");
                continue;
            }
            else
            {
                sort(a,a+m);
                if(a[0]==1||a[m-1]==n)
                {
                    printf("NO\n");
                    continue;
                }
                else
                {
                    int cont=0;
                    int flag=1;
                    for(int i=2;i<m;i++)
                    {
                        if(a[i]-a[i-2]<=2)
                        {
                            printf("NO\n");
                            flag=0;
                            break;
                        }
                    }
                    if(flag==1)
                        printf("YES\n");
                }
            }
        }
        return 0;
    }

[Link 1]: http://codeforces.com/problemset/problem/362/B