Bear and Three Balls

B - Bear and Three Balls（sort）
Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u
Submit Status

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, 5and 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

Input

4
18 55 16 17

Output

YES

Input

6
40 41 43 44 44 44

Output

NO

Input

8
5 972 3 4 1 4 970 971

Output

YES

题意很简单找到三个重复的数后去重就好了
先排序再去重

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
int ball[10000];
int b[1000];
int main(){
while(~scanf("%d",&n)){
memset(ball,0,sizeof(ball));
memset(b,0,sizeof(b));
int size;
for(int i=0;i<n;i++)
scanf("%d",&b[i]);

sort(b,b+n);//排序
int k=0;
for(int i=0;i<n;i++){
if(b[i]!=b[i+1]){
ball[k]=b[i];k++;}
}//去重
int y=0;
for(int i=0;i<k-2;i++){
if(ball[i]+1==ball[i+1]&&ball[i]+2==ball[i+2]){
y=1;
break;
}
}//判断有没有三个连续数
if(y)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}