HDU

Numbers

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

 

Source

2017 Multi-University Training Contest - Team 9

 

Recommend

liuyiding

 

题意：给出一串数字C，C是由数列A，B组成的。B是A中两两的和。输出数列A。

解题思路：先对C排序，然后最小的那个数字肯定是数列A，先把最小的加入到答案列表A。然后对于每一个次小的与A中每个数字做和，并从给定的数字中去掉，循环以上过程。

附上官方题解

1008

将b数组排序，取出最小的两项作为a_1,a_2a​1​​,a​2​​，删除a_1,a_2,a_1+a_2a​1​​,a​2​​,a​1​​+a​2​​，再取出最小项作为a_3a​3​​，再删除a_3,a_1+a_3,a_2+a_3a​3​​,a​1​​+a​3​​,a​2​​+a​3​​，再取出最小项作为a_4a​4​​,依次列推。

#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 200005using 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;}