CF IndiaHack A 水题

题目连接

http://codeforces.com/contest/653/problem/A

Description

Limak is a little polar bear. He has n balls, the i-th ball has size ti.

Limak wants to give one ball to each of his three friends. Giving gifts isn’t easy — there are two rules Limak must obey to make friends happy

* No two friends can get balls of the same size.
* No two friends can get balls of sizes that differ by more than 2

For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can’t choose balls with sizes 5, 5 and 6 (two friends would get balls of the same size), and he can’t choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).

Your task is to check whether Limak can choose three balls that satisfy conditions above.

Input

The first line of the input contains one integer n (3 ≤ n ≤ 50) — the number of balls Limak has.

The second line contains n integers t1, t2, …, tn (1 ≤ ti ≤ 1000) where ti denotes the size of the i-th ball.

Output

Print “YES” (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print “NO” (without quotes).

Sample Input

4

18 55 16 17

Sample Output

YES

题意

能否找到三个不同的数，他们任意两个数的差都是1

代码

#include<bits/stdc++.h>
typedef long long ll;
#define maxn 20010
#define INF  1e9+7
using namespace std;
int a[110];
int main(int argc, char const *argv[])
{
int n;
cin>>n;
for (int i = 0; i < n; ++i)
{
cin>>a[i];
}
int m=1;
int i=0;
sort(a,a+n);
while(i!=n)
{
if (m==3)
{
printf("Yes\n");
return 0;
}
if (a[i]==a[i+1])
{
i++;
continue;
}
if(a[i+1]-a[i]!=1){
m=1;
i++;
continue;
}
else{
m++;
i++;
}
}
printf("No\n");
return 0;
}

Ps

需要注意的是，当值相等的时候m的计数不是重新开始，而是跳过这个数向下计数