hdu 6168 模拟+优先队列

zk has n numbers  a1,a2,...,an. For each (i,j) satisfying 1≤i (ai+aj). These new numbers could make up a new sequence  b1，b2,...,bn(n−1)/2.
LsF wants to make some trouble. While zk is sleeping, Lsf mixed up sequence a and b with random order so that zk can't figure out which numbers were in a or b. "I'm angry!", says zk.
Can you help zk find out which n numbers were originally in a?

 

Input

Multiple test cases(not exceed 10).
For each test case:
∙The first line is an integer m(0≤m≤125250), indicating the total length of a and b. It's guaranteed m can be formed as n(n+1)/2.
∙The second line contains m numbers, indicating the mixed sequence of a and b.
Each  ai is in [1,10^9]

 

Output

For each test case, output two lines.
The first line is an integer n, indicating the length of sequence a;
The second line should contain n space-seprated integers  a1,a2,...,an(a1≤a2≤...≤an). These are numbers in sequence a.
It's guaranteed that there is only one solution for each case.

 

Sample Input

6 2 2 2 4 4 4 21 1 2 3 3 4 4 5 5 5 6 6 6 7 7 7 8 8 9 9 10 11

 

Sample Output

3 2 2 2 6 1 2 3 4 5 6

 

Source

2017 Multi-University Training Contest - Team 9

 

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首先最小的两个一定是a数组里边的然后把a【0】+a【1】加入到b数组（优先队列）中，在找下一个如果等于b数组中的数就继续，如果不等于b数组里的数就是a数组里的数，更新b数组就可以了

ac代码：

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
int a[125255];
int c[200001];
int b[200001];
int main()
{
    int m;
    while(cin>>m)
    {
        priority_queue,greater >q1;
        while(!q1.empty())
        q1.pop();
        for(int i=0; i