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.
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.
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).
4
18 55 16 17
YES
6
40 41 43 44 44 44
NO
8
5 972 3 4 1 4 970 971
YES
<span style="font-size:18px;">#include <iostream>
#include <algorithm>
#include <set>
#include <cstdio>
#include <cstring>
using namespace std;
set <int> s;
set <int> ::iterator it;
int a[105];
int main()
{
int n, x;
while(cin>>n)
{
for(int i=0; i<n; i++)
{
cin>>x;
s.insert(x);
}
memset(a, 0, sizeof(a));
int cnt = 0;
for(it=s.begin(); it!=s.end(); it++)
a[cnt++] = *it;///每个都存一遍 实在是没招了 set学的不好呀。。。
bool ok = 0;
for(int i=1; i<cnt; i++)
{
if(a[i]==a[i-1]+1 && a[i]+1==a[i+1])
{
ok = 1;
break;
}
}
if(ok)
puts("YES");
else
puts("NO");
}
return 0;
}</span>
接下来我要说的就是这篇博客的重点了（有没有想到高中老师呀...）介绍一个 STL 算法，unique()，这个函数是用来去重的，这还是我一队友告诉我的呢，我感觉自己好low啊，我详细的说一下：
<span style="font-size:18px;">#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
int a[105];
int main()
{
int n, x;
while(cin>>n)
{
for(int i=0; i<n; i++)
cin>>a[i];
sort(a,a+n);
int tmp = unique(a,a+n)-a;///头文件 <algorithm>，还是挺好的，去重用的
bool ok = 0;
for(int i=1; i<tmp; i++)
{
if(a[i]==a[i-1]+1 && a[i]+1==a[i+1])
{
ok = 1;
break;
}
}
if(ok)
puts("YES");
else
puts("NO");
}
return 0;
}
</span>
