codeforces IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2) A. Bear and Three Balls
A. Bear and Three Balls
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Limak is a little polar bear. He has n balls, the i-th ball hassize ti.
Limak wants togive 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 morethan 2.
For example,Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can'tchoose balls with sizes 5, 5 and6 (two friends would get balls of the same size), andhe can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ bymore than 2).
Your task is tocheck whether Limak can choose three balls that satisfy conditions above.
Input
The first lineof the input contains one integer n (3 ≤ n ≤ 50) — thenumber of balls Limak has.
The second linecontains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1000) where ti denotesthe size of the i-th ball.
Output
Print "YES" (without quotes) if Limak canchoose three balls of distinct sizes, such that any two of them differ by nomore than 2. Otherwise, print "NO" (without quotes).
Examples
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
Note
In the first sample, there are 4 balls and Limak is able to choose three of themto satisfy the rules. He must must choose balls with sizes18, 16 and 17.
In the second sample, there is no way to give gifts tothree friends without breaking the rules.
In the third sample, there is even more than one wayto choose balls:
1. Choose balls with sizes 3, 4 and 5.
2. Choose balls with sizes 972, 970, 971.
这道题就是判断给定的一串数字中是否存在三个连续的整数；
对这些数排序后，在进行判断；
注意存在相等的数时，要接着向后在移一位；
#include<bits/stdc++.h>
using namespace std;
int num[51];
int main()
{
int n;
scanf("%d",&n);
for(int i=0;i<n;++i)
{
scanf("%d",&num[i]);
}
sort(num,num+n);
boolflag=0;
for(int i=0;i<n-2;++i)
{
for(int j=i+1;j<n-1;++j)
{
if(num[j]-num[i]>1)
break;
for(int k=j+1;k<n;++k)
{
if(num[k]-num[j]>1) break;
if(num[k]-num[j]==1&&num[j]-num[i]==1)
{
flag=1;
break;
}
}
if(flag) break;
}
if(flag) break;
}
if(flag)cout<<"YES"<<endl;
else cout<<"NO"<<endl;
return 0;
}