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IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2)——A - Bear and Three Balls(unique函数的使用)...
阅读量:5223 次
发布时间:2019-06-14

本文共 2493 字,大约阅读时间需要 8 分钟。

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 45 and 3, or balls with sizes 9091 and 92. But he can't choose balls with sizes 55and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 3031 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 sizes1816 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 34 and 5.
  2. Choose balls with sizes 972970971.

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
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
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
>list[i]; } sort(list,list+n); int *p=unique(list,list+n); int k=p-list; for (i=0; i

转载于:https://www.cnblogs.com/Blackops/p/5356391.html

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