HDU6168-Numbers

Numbers

                                                                   Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
                                                                                             Total Submission(s): 348    Accepted Submission(s): 175

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

  

   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

 

题意：给你两个序列A,B，序列B中的数是A中任意两个数的和。现在给出你A,B序列混在一起的数，让你找出A序列输出

解题思路：两个优先队列随便搞一搞就好了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <map>
#include <set>
#include <algorithm>
#include <vector>
#include <bitset>
#include <stack>
#include <queue>
#include <unordered_map>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int m;
LL a;
LL ans[1000009];

struct node
{
    LL x;
    bool operator<(const node &a)const
    {
        return x>a.x;
    }
} pre,pre1,nt;

int main()
{
    while(~scanf("%d",&m))
    {
        if(m==0) {printf("0\n");continue;}
        priority_queue<node>q1,q2;
        for(int i=1; i<=m; i++)
        {
            scanf("%lld",&a);
            pre.x=a;
            q1.push(pre);
        }
        if(m==1) {printf("1\n%lld\n",pre.x);continue;}
        int cnt=1;
        for(int i=0; i<2; i++)
        {
            pre=q1.top();
            q1.pop();
            ans[cnt++]=pre.x;
        }
        pre.x=ans[1]+ans[2];
        q2.push(pre);
        while(!q1.empty())
        {
            pre=q1.top();
            q1.pop();
            if(!q2.empty()) pre1=q2.top();
            if(!q2.empty()&&pre.x==pre1.x) {q2.pop();continue;}
            for(int i=1;i<cnt;i++)
            {
                nt.x=ans[i]+pre.x;
                q2.push(nt);
            }
            ans[cnt++]=pre.x;
        }
        int flag=1;
        printf("%d\n",cnt-1);
        for(int i=1;i<cnt;i++)
        {
            if(!flag) printf(" ");
            else flag=0;
            printf("%lld",ans[i]);
        }
        printf("\n");
    }
    return 0;
}