【IndiaHacks 2016 - Online Edition (Div 1 + Div 2) ErrichtoA】【水题】Bear and Three Balls 是否有数值相差1的三个数

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 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).

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 them to satisfy the rules. He must must choose balls with sizes18, 16 and 17.
In the second sample, there is no way to give gifts to three friends without breaking the rules.
In the third sample, there is even more than one way to choose balls:

Choose balls with sizes 3, 4 and 5.

Choose balls with sizes 972, 970, 971.

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int n;
int a[1010];
bool check()
{
for (int i = 2; i < 1000; ++i)
{
if (a[i - 1] && a[i] && a[i + 1])return 1;
}
return 0;
}
int main()
{
while (~scanf("%d", &n))
{
for (int i = 1; i <= n; ++i)
{
int x;
scanf("%d", &x);
a[x] = 1;
}
puts(check() ? "YES" : "NO");
}
return 0;
}
/*
【trick&&吐槽】
>_<判断是否有三个连续数值的数，可不能用排序来做啊

【题意】
有n（50）个数t[]，每个数的数值范围都为[1,1000]
让你求出是否有x,x+1和x+2这样的三个数

【类型】
水题

【分析】
计数排序是你最好的选择

【时间复杂度&&优化】
O（1000）

*/