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 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:

  1. Choose balls with sizes 3, 4 and 5.
  2. Choose balls with sizes 972, 970, 971.

题目本身不难,就是问你有没有三个连续的数字,比如1 2 3,45 46 47这样的,如果你用等差数列去判断,那存在一个坑点就是1 2 2 3,这样不是等差数列,但是也是可行的即输出YES。因此用了unique函数,这个函数的作用是把一个排序后的数组合并连续区间内相同的数字,比如1 2 2 2 3会变成1 2 3,那么剩下的两个2去哪了呢,被移到了数组的末尾。此函数本身返回一个完成合并区间的超尾指针,跟.end()差不多。因此使用之后的数组大小应该是N=unique(list,list+n)-list;

代码:

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
int list[100];
int main (void)
{
int n,i,j,k;
while (cin>>n)
{
memset(list,0,sizeof(list));
bool flag=0;
for (i=0; i<n; i++)
{
cin>>list[i];
}
sort(list,list+n);
int *p=unique(list,list+n);
int k=p-list;
for (i=0; i<k-2; i++)
{
if(list[i]+1==list[i+1]&&list[i+1]+1==list[i+2]&&list[i]+2==list[i+2])
{
flag=true;
break;
}
}
if(flag)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}

IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2)——A - Bear and Three Balls(unique函数的使用)的更多相关文章

  1. IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2) B. Bear and Compressing

    B. Bear and Compressing 题目链接  Problem - B - Codeforces   Limak is a little polar bear. Polar bears h ...

  2. IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2) E - Bear and Forgotten Tree 2 链表

    E - Bear and Forgotten Tree 2 思路:先不考虑1这个点,求有多少个连通块,每个连通块里有多少个点能和1连,这样就能确定1的度数的上下界. 求连通块用链表维护. #inclu ...

  3. IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2) E. Bear and Forgotten Tree 2 bfs set 反图的生成树

    E. Bear and Forgotten Tree 2 题目连接: http://www.codeforces.com/contest/653/problem/E Description A tre ...

  4. IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2) D. Delivery Bears 二分+网络流

    D. Delivery Bears 题目连接: http://www.codeforces.com/contest/653/problem/D Description Niwel is a littl ...

  5. IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2) C. Bear and Up-Down 暴力

    C. Bear and Up-Down 题目连接: http://www.codeforces.com/contest/653/problem/C Description The life goes ...

  6. IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2) B. Bear and Compressing 暴力

    B. Bear and Compressing 题目连接: http://www.codeforces.com/contest/653/problem/B Description Limak is a ...

  7. IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2) A. Bear and Three Balls 水题

    A. Bear and Three Balls 题目连接: http://www.codeforces.com/contest/653/problem/A Description Limak is a ...

  8. CodeForces 653 A. Bear and Three Balls——(IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2))

    传送门 A. Bear and Three Balls time limit per test 2 seconds memory limit per test 256 megabytes input ...

  9. IndiaHacks 2016 - Online Edition (CF) . D

    这题思路很简单,二分m,求最大流是否大于等于x. 但是比赛过程中大部分的代码都被hack了... 精度问题,和流量可能超int 关于精度问题,这题真是提醒的到位,如果是先用二分将精度控制在10^-8左 ...

随机推荐

  1. 使用com.sun.imageio.plugins.png.PNGMetadata读取图片的元数据

    所谓图片元数据,就是除了我们肉眼看到的图片内容外,隐藏在这些内容背后的一些技术数据. 本文介绍如何使用Java代码将一张图片的隐藏信息读取出来. 首先不需要下载任何额外的Java库,用JDK自带的库就 ...

  2. 日常-acm-鸡兔同笼

    已知鸡和兔总数量n,总腿数m.输入n和m,依次输出鸡的数量和兔的数量.如果无解,则输出No answer. 样例输入: 14 32 样例输出: 12 2 样例输入: 10 16 样例输出: No an ...

  3. 学习Unity 4.6新GUI系统

    (搬运自我在SegmentFault的博客) 最近在学习Unity的过程中,自己做一款小游戏自娱自乐.自然需要用到GUI.但4.5中的GUI很难用,一个选择是传说中的NGUI插件.但对于4.6中的新G ...

  4. A winner is a dreamer who never gives up

    A winner is a dreamer who never gives up. 成功者是坚持梦想不放弃的人.(Nelson Mandela)

  5. .net代码获取节点以及读取属性

    获取配置文件的节点,可以使用System.Configuration.ConfigurationManager.GetSection方法获取指定的节点,以sessionstate节点为例,如果需要获取 ...

  6. 数据库要素 ER

    数据库的要素即为ER: 即为表和关系. 再往下即为字段.记录. 往上即为数据操作.管理: 包含多表操作: 在往上为事务. 再往上为大数据.高并发.

  7. LeetCode分类-前400题

    1. Array 基础 27 Remove Element 26 Remove Duplicates from Sorted Array 80 Remove Duplicates from Sorte ...

  8. Dojo操作dom元素的样式

    1.使用dom-style的set方法,可以直接设置dom元素的样式属性,这和使用dom元素的style属性效果一样. 2.使用dom-class的replace方法可以替换某个dom元素的样式,ad ...

  9. 监控电脑CPU,内存,文件大小,硬盘空间,IP,用户名

    public class MonitorTools { /// <summary> /// 获取具体进程的内存,线程等参数情况 /// </summary> /// <p ...

  10. linux系统下的用户文件句柄数限制

    linux系统下的用户文件句柄数限制 文章来源:企鹅号 为什么要修改用户打开的文件数 系统默认单个进程可以打开1024个文件,对于一些应用如tomcat.oracle等,运行时经常open成千上万个文 ...