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+;
int main()
{
int n;
while(cin>>n)
{
ll a[maxn];
bool flag= false,flag1=false;
for(int i=;i<n;i++)
{
cin>>a[i];
if(a[i]&)
flag=true;
else
flag1=true; }
if(flag&&flag1)
sort(a,a+n);
copy(a,a+n,ostream_iterator<ll>(cout," "));
cout<<endl;
}
return ;
}
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