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).
18 55 16 17
40 41 43 44 44 44
5 972 3 4 1 4 970 971
就是给定 n 个数，让你求的就是是否存在三个连续的数。。。
这个题本来这么简单不想写的，但是突然发现一个函数还是比较不错的，unique()，就是这个函数，本来我是打算暴力的后来一寻思太麻烦了，就用集合写的，写到一半突然发现迭代器 it 不能加2，所以炸了，后来就用一个数组存的每一个*it 的数，有点小麻烦，具体看我第一个代码
My First AC Code：
<span style="font-size:18px;">#include <iostream>
using namespace std;
set <int> s;
set <int> ::iterator it;
int n, x;
for(int i=0; i<n; i++)
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;