zoukankan      html  css  js  c++  java
  • Codeforces Round #371 (Div. 2) B. Filya and Homework 水题

    B. Filya and Homework

    题目连接:

    http://codeforces.com/contest/714/problem/B

    Description

    Today, hedgehog Filya went to school for the very first time! Teacher gave him a homework which Filya was unable to complete without your help.

    Filya is given an array of non-negative integers a1, a2, ..., an. First, he pick an integer x and then he adds x to some elements of the array (no more than once), subtract x from some other elements (also, no more than once) and do no change other elements. He wants all elements of the array to be equal.

    Now he wonders if it's possible to pick such integer x and change some elements of the array using this x in order to make all elements equal.

    Input

    The first line of the input contains an integer n (1 ≤ n ≤ 100 000) — the number of integers in the Filya's array. The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109) — elements of the array.

    Output

    If it's impossible to make all elements of the array equal using the process given in the problem statement, then print "NO" (without quotes) in the only line of the output. Otherwise print "YES" (without quotes).

    Sample Input

    5
    1 3 3 2 1

    Sample Output

    YES

    Hint

    题意

    给你n个数,对于每一个数,你得选择加上x还是减去x,问你最后所有数是否能够都相等。

    题解:

    千万不要读错题……

    然后读懂之后,排序去重瞎判断一下就好了。

    代码

    #include<bits/stdc++.h>
    using namespace std;
    const int maxn = 1e5+6;
    int n,a[maxn];
    int main()
    {
        scanf("%d",&n);
        vector<int>A;
        for(int i=1;i<=n;i++)
        {
            int x;scanf("%d",&x);
            A.push_back(x);
        }
        sort(A.begin(),A.end());
        A.erase(unique(A.begin(),A.end()),A.end());
        if(A.size()<=2)
            printf("YES
    ");
        else if(A.size()==3&&A[2]+A[0]==A[1]*2)
            printf("YES
    ");
        else
            printf("NO
    ");
    }
  • 相关阅读:
    4-数组、指针与字符串1.3-this指针
    Linux命令----cp
    Linux命令----mv
    Linux命令----rm
    PHP7下的协程实现 转
    php生成器 yield 转
    python并发编程之多进程(实践篇) 转
    python 多进程
    线程创建 线程数
    多任务 执行
  • 原文地址:https://www.cnblogs.com/qscqesze/p/5874492.html
Copyright © 2011-2022 走看看