HDU - 6168 Numbers

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

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

 

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 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;
}