hdu 6168 Numbers

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6168

Numbers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 631    Accepted Submission(s): 335


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

 

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解析：求原来的数组，先排下序，最小的两个一定是原数组，然后一个一个的取，取完之后进行组合，把取出来的和求的和从数组中删除，一直重复这个操作即可

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 125250+100;

int a
, b
;
map<int, int> mp;

int main()
{
int n, len;
while(~scanf("%d", &n))
{
mp.clear();
len = 0;
for(int i = 0; i < n; i++)
{
scanf("%d", &a[i]);
mp[a[i]]++;
}
sort(a, a+n);
b[len++] = a[0]; b[len++] = a[1];
mp[a[0]]--; mp[a[1]]--;
int cur = a[0] + a[1], j = 2;
mp[cur]--;
while(j < n)
{
while(j < n && !mp[a[j]]) j++;
if(j >= n) break;
int cur = a[j];
mp[cur]--;
for(int i = 0; i < len; i++)
{
int d = b[i] + cur;
if(mp[d]) mp[d]--;
}
b[len++] = cur;
}
printf("%d\n", len);
for(int i = 0; i < len; i++) printf("%d%c", b[i], i == len-1 ? '\n' : ' ');
}
return 0;
}