Vitaly has an array of n distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
Help Vitaly. Divide the given array.
The first line of the input contains integer n (3?≤?n?≤?100). The second line contains n space-separated distinct integers a1,?a2,?...,?an (|ai|?≤?103) — the array elements.
In the first line print integer n1 (n1?>?0) — the number of elements in the first set. Then print n1 numbers — the elements that got to the first set.
In the next line print integer n2 (n2?>?0) — the number of elements in the second set. Then print n2 numbers — the elements that got to the second set.
In the next line print integer n3 (n3?>?0) — the number of elements in the third set. Then print n3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
3 -1 2 0
1 -1 1 2 1 0
程序好长，但是思想很简单，就分好类就可以了，而且符合条件的就可以任意分类了。
#include <vector>
#include <string>
#include <iostream>
#include <stdio.h>
using namespace std;
void ArrayDivideSub()
{
int n, a;
cin>>n;
vector<int> lesZero;
vector<int> larZero;
vector<int> equZero;
for (unsigned i = 0; i < n; i++)
{
scanf("%d", &a);
if (a < 0) lesZero.push_back(a);
else if (a > 0) larZero.push_back(a);
else equZero.push_back(a);
}
if (larZero.empty())
{
if (lesZero.size())
{
larZero.push_back(lesZero.back());
lesZero.pop_back();
}
if (lesZero.size())
{
larZero.push_back(lesZero.back());
lesZero.pop_back();
}
}
if (lesZero.size() && lesZero.size() % 2 == 0)
{
equZero.push_back(lesZero.back());
lesZero.pop_back();
}
cout<<lesZero.size()<<‘ ‘;
for (unsigned i = 0; i < lesZero.size(); i++)
{
cout<<lesZero[i]<<‘ ‘;
}
cout<<endl;
cout<<larZero.size()<<‘ ‘;
for (unsigned i = 0; i < larZero.size(); i++)
{
cout<<larZero[i]<<‘ ‘;
}
cout<<endl;
cout<<equZero.size()<<‘ ‘;
for (unsigned i = 0; i < equZero.size(); i++)
{
cout<<equZero[i]<<‘ ‘;
}
}
codeforces A. Array题解,布布扣,bubuko.com
原文：http://blog.csdn.net/kenden23/article/details/24878079