HDU6168

Numbers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2015    Accepted Submission(s): 710

Problem Description

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?

 

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

62 2 2 4 4 4211 2 3 3 4 4 5 5 5 6 6 6 7 7 7 8 8 9 9 10 11

 

Sample Output

32 2 261 2 3 4 5 6
#include<iostream>  #include<deque>  #include<memory.h>  #include<stdio.h>  #include<map>  #include<string>  #include<algorithm>  #include<vector>  #include<math.h>  #include<stack>  #include<queue>  #include<set>  #define inf 1073741823  #define MAXV 200005  using namespace std;    int m;  int temp;    int main(){        while(~scanf("%d",&m)){          map<int,int> num;//记录数字出现过多少次          vector<int> mynum;//数列C          vector<int> ans;//数列A          for(int i=0;i<m;i++){              scanf("%d",&temp);              mynum.push_back(temp);              if(num[temp]==0)                  num[temp]=1;              else                  num[temp]++;            }            if(m==0)//特殊处理          {              cout<<0<<endl;              continue;          }              sort(mynum.begin(),mynum.end());          ans.push_back(mynum[0]);            for(int i=1;i<mynum.size();i++){              if(num[mynum[i]]==0)//如果用完了，下一个                  continue;                for(int j=0;j<ans.size();j++)                  num[ans[j]+mynum[i]]--;                ans.push_back(mynum[i]);              num[mynum[i]]--;            }              printf("%d\n",ans.size());          for(int i=0;i<ans.size();i++)              if(i==ans.size()-1)              printf("%d\n",ans[i]);          else                  printf("%d ",ans[i]);        }      return 0;  }