zk has n numbers a1,a2,...,ana1,a2,...,an. For each (i,j) satisfying 1≤i<j≤n, zk generates a new number (ai+aj)(ai+aj). These new numbers could make up a new sequence b1，b2,...,bn(n−1)/2b1，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 aiai 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)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
这个看代码吧。QAQ
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int n,m;
vector <int> a,b,c,ans;
map <int,int> mmp;
int main()
{
while(~scanf("%d",&n))
{
a.clear();
b.clear();
c.clear();
ans.clear();
mmp.clear();
int temp;
for(int i=0;i<n;i++)
{
scanf("%d",&temp);
a.push_back(temp);
if(mmp[temp]==0) mmp[temp]=1;
else mmp[temp]++;
}
sort(a.begin(),a.end());
ans.push_back(a[0]);
for(int i=1;i<n;i++)
{
if(mmp[a[i]]==0) continue;
for(int j=0;j<ans.size();j++)
{
mmp[ans[j]+a[i]]--;
}
ans.push_back(a[i]);
mmp[a[i]]--;
}
printf("%d\n",ans.size());
vector <int> ::iterator it;
for(it=ans.begin();it!=ans.end();it++)
{
if(it==ans.begin()) cout<<*it;
else cout<<" "<<*it;
}
cout<<endl;
}
return 0;
}