题目链接:
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
418 55 16 17
output
YES
input
640 41 43 44 44 44
output
NO
input
85 972 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.
题意:就是问你是否有连续的三个数;
AC代码:
#include <bits/stdc++.h> using namespace std; int a[1000],num[1005]; int main() { int n; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&a[i]); num[a[i]]++; } for(int i=1;i<=1000;i++) { if(num[i]&&num[i+1]&&num[i+2]) { cout<<"YES"; return 0; } } cout<<"NO"; return 0; }