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:
    HTTP 常见状态码
    SpringMVC 入门
    Maven 整合SSH框架
    SSH 框架整合总结
    Maven 整合SSH框架之pom.xml
    Maven 入门
    WebService 综合案例
    CXF 框架
    jQuery高级
    JavaScript补充:BOM(浏览器对象模型)
  • 原文地址:https://www.cnblogs.com/qscqesze/p/5874492.html
Copyright © 2011-2022 走看看