题目链接：http://codeforces.com/contest/1174/problem/A
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
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。
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+10; int main() { int n; while(cin>>n) { int a[maxn]; int x; cin>>x; a[0]=x; int flag=a[0]; int num=1; for(int i=1;i<2*n;i++) { cin>>a[i]; if(a[i]==flag) num++; } if(num==2*n) cout<<"-1"<<endl; else { int l=accumulate(a,a+n,0); cerr<<"l="<<l<<endl;//cerr对提交无影响 int r=accumulate(a+n,a+2*n,0); cerr<<"r="<<r<<endl; if(l!=r) { for(int i=0;i<2*n;i++) { cout<<a[i]<<‘ ‘; } cout<<endl; } else { sort(a,a+2*n); for(int i=0;i<2*n;i++) cout<<a[i]<<‘ ‘; cout<<endl; } } } return 0; }
Codeforces Round #563 (Div. 2)A
原文：https://www.cnblogs.com/Vampire6/p/10989800.html