zoukankan      html  css  js  c++  java
  • Codeforces Round #563 (Div. 2)B

    B.Ehab Is an Odd Person

    题目链接:http://codeforces.com/contest/1174/problem/B

    题目

    You’re given an array a of length n. You can perform the following operation on it as many times as you want:

    Pick two integers i and j (1≤i,j≤n) such that ai+aj is odd, then swap ai and aj.

    What is lexicographically the smallest array you can obtain?

    An array x is lexicographically smaller than an array y if there exists an index i such that xi<yi, and xj=yj for all 1≤j<i. Less formally, at the first index i in which they differ, xi<yi

    Input
    The first line contains an integer n
    (1≤n≤105) — the number of elements in the array a.
    The second line contains n space-separated integers a1, a2, …, an (1≤ai≤109) — the elements of the array a.

    Output
    The only line contains n
    space-separated integers, the lexicographically smallest array you can obtain.

    Example

    input

    3

    4 1 7

    output

    1 4 7

    题意

    给你一个长度为a的数组,选择两个数之和为奇数,可以交换这两个数,要使得这个数组字典序最小,输出变换后的这个数组。

    思路

    计算奇数和偶数的个数,一旦只有奇数或者只有偶数,则不存在两个数之和为奇数,故直接将原来数组输出就行,

    只要都存在奇数或者偶数,则可以通过各种调法调成单调递增的数组,所以只要将其排序后输出即可。

    //
    // Created by hjy on 19-6-4.
    //
    
    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    const int maxn=2e5+10;
    int main()
    {
        int n;
        while(cin>>n)
        {
            ll a[maxn];
            bool flag= false,flag1=false;
            for(int i=0;i<n;i++)
            {
                cin>>a[i];
                if(a[i]&1)
                    flag=true;
                else
                    flag1=true;
    
            }
            if(flag&&flag1)
                sort(a,a+n);
            copy(a,a+n,ostream_iterator<ll>(cout," "));
            cout<<endl;
        }
        return 0;
    }
  • 相关阅读:
    MD支持新标签跳转
    线上问题cpu100处理记录
    OpenShift 4.6方式下OperatorHub的变化
    OpenShift 4.5.7 版本基础镜像下载
    GLPI企业使用(一),连接AD域,LDAP登录。
    GLPI配置文件说明:默认权限组
    企业服务器规划
    港股通转托管
    mui实现下拉刷新以及click事件无法响应问题
    asp.net core+websocket实现实时通信
  • 原文地址:https://www.cnblogs.com/Vampire6/p/10989806.html
Copyright © 2011-2022 走看看