Codeforces Round #563 (Div. 2)A

A. Ehab Fails to Be Thanos

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

题目

You’re given an array a of length 2n. Is it possible to reorder it in such way so that the sum of the first n elements isn’t equal to the sum of the last n elements?

Input
The first line contains an integer n(1≤n≤1000), where 2n is the number of elements in the array a.
The second line contains 2n space-separated integers a1, a2, …, a2n (1≤ai≤106) — the elements of the array a

Output
If there’s no solution, print “-1” (without quotes). Otherwise, print a single line containing 2n
space-separated integers. They must form a reordering of a. You are allowed to not change the order

Example

input3

1 2 2 1 3 1

output

2 1 3 1 1 2

题意

给你一个长度为2n的数组a，对其重新排序，使得前n和后n元素的总和不想等，若无法排序输出-1，可以则输出重新排序过的长度为2n的数组a。

思路

先判断a数组中的数是否一样，若一样则无法排序使其前n个数之和不为后n个数之和，输出-1；

否则，求出前n与后n个数的和，判断他们是否相等，如果相等，将a排序再顺序输出即可，不想等则直接将原来a数组反向输出即可。

//
// 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)
    {
        int a[maxn];
        int x;
        cin>>x;
        a[]=x;
        int flag=a[];
        int num=;
        for(int i=;i<*n;i++)
        {
            cin>>a[i];
            if(a[i]==flag)
                num++;
        }
        if(num==*n)
            cout<<"-1"<<endl;
        else
        {
            int l=accumulate(a,a+n,);
            cerr<<"l="<<l<<endl;//cerr对提交无影响

            int r=accumulate(a+n,a+*n,);
            cerr<<"r="<<r<<endl;
            if(l!=r) {
                for(int i=;i<*n;i++)
                {
                    cout<<a[i]<<' ';
                }
                cout<<endl;
            } else
            {
                sort(a,a+*n);
                for(int i=;i<*n;i++)
                    cout<<a[i]<<' ';
                cout<<endl;
            }

        }

    }
    return ;
}