HDU 6168 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

题意：

给出数组C 数组C由数组A、B组成

数组B由数组A两两相加所得

求原数组A

题解：

首先排个序 一定可以确定最小的两个元素是原数组A

因为没有任何数字相加可以得到这两个最小的元素

齐次 用这两个最小的元素丢进multiset中 erase相加所得的元素

然后再用齐次小的两个元素

#include<stdio.h>
#include<string.h>
#include<set>
#define ll long long
using namespace std;
const int MAX = 1e6;
int ans[MAX];

int main(){
int m;
while(~scanf("%d",&m)){
multiset<int>s;
for(int i = 0;i < m;i++){
int x;
scanf("%d",&x);
s.insert(x);
}
int cnt = 0;
ans[cnt++] = *s.begin();
s.erase(s.begin());
while(!s.empty()){
ans[cnt++] = *s.begin();
s.erase(s.begin());
for(int i = 0;i < cnt - 1;i++){
s.erase(s.lower_bound(ans[cnt-1] + ans[i]));
}
}
printf("%d\n",cnt);
for(int i = 0;i < cnt;i++){
if(i)
printf(" %d",ans[i]);
else
printf("%d",ans[i]);
}
puts("");
}
return 0;
}