CF——A. Ehab Fails to Be Thanos（第一场cf爆零之战）

最新推荐文章于 2019-07-23 11:57:01 发布

原创最新推荐文章于 2019-07-23 11:57:01 发布 · 416 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#奇怪输出 #水题

CF 专栏收录该内容

5 篇文章

订阅专栏

本文解析了一场Codeforces比赛中的题目A.EhabFailstoBeThanos，介绍了如何判断并输出一个2n长度序列的任意排列，使得前n项和与后n项和不相等的方法。

A. Ehab Fails to Be Thanos

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.
Examples
input
3
1 2 2 1 3 1
output
2 1 3 1 1 2
input
1
1 1
output
-1
Note
In the first example, the first n elements have sum 2+1+3=6 while the last n elements have sum 1+1+2=4. The sums aren’t equal.

In the second example, there’s no solution.
ps:打的第一场cf，由于这道题始终没搞懂题目意思，不懂怎样输出，所以爆零了。最后看了别人代码才才知道，还有这种输出的格式，长知识了。题意就是给一个2n的序列，任意交换位置，使前n位之和与后n位的和不等，如果满足情况，则输出任意一列情况 就可以了，否则输出-1。哎，菜是原罪，请原谅我孤陋寡闻

#include<bits/stdc++.h>
using namespace std;
int arr[10005];
int main(){
    int n;
    cin>>n;
    for(int i=0;i<2*n;i++){
        cin>>arr[i];
    }
    sort(arr,arr+2*n);
    int suma=0,sumb=0;
    for(int i=0;i<n;i++){
        suma+=arr[i];
    }
    for(int i=n;i<2*n;i++){
        sumb+=arr[i];
    }
    if(suma==sumb){
        cout<<"-1"<<endl;
    }else{
        for(int i=0;i<2*n;i++){
            cout<<arr[i]<<" ";
        }
        cout<<endl;
    }
    return 0;
}