hdu 6168 模拟+优先队列

zk has n numbers  a1,a2,...,an . For each (i,j) satisfying 1≤i<j≤n, zk generates a new number  (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?

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]

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.

  
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

  
3
2 2 2
6
1 2 3 4 5 6

2017 Multi-University Training Contest - Team 9

liuyiding   |   We have carefully selected several similar problems for you:   6170  6169  6168  6167  6166 

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 <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
int a[125255];
int c[200001];
int b[200001];
int main()
{int m;while(cin>>m){priority_queue<int,vector<int>,greater<int> >q1;while(!q1.empty())q1.pop();for(int i=0; i<m; i++)scanf("%d",&c[i]);sort(c,c+m);if(m==0){printf("0\n");printf("\n");}else if(m==1){printf("1\n");printf("%d\n",c[0]);}else{int lena=0;int k=0;int n=sqrt(2*m);a[lena++]=c[0];a[lena++]=c[1];int ph=c[0]+c[1];q1.push(ph);for(int i=2; i<m; i++){if(!q1.empty()&&c[i]==q1.top()){q1.pop();continue;}else{a[lena++]=c[i];for(int j=0; j<lena-1; j++){q1.push(a[lena-1]+a[j]);//cout<<a[lena-1]+a[j]<<endl;}}if(lena==n)break;}printf("%d\n",n);for(int i=0; i<lena-1; i++)printf("%d ",a[i]);printf("%d\n",a[lena-1]);}}return 0;
}