Numbers

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

Total Submission(s): 356    Accepted Submission(s): 180

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

题意：

给你两个序列A,B，序列B中的数是A中任意两个数的和。现在给出你A,B序列混在一起的数，让你找出A序列输出。

思路：

每次加进去一个筛掉后最小的数，然后与前面已经有的数把剩下的数筛一下就好。

#include 
    
     #define INF 0x3f3f3f3f#define ms(x,y) memset(x,y,sizeof(x))using namespace std;typedef long long ll;const double pi = acos(-1.0);const int mod = 1e9 + 7;const int maxn = 125350;int a[maxn], ans[maxn];int main(){	//freopen("in.txt","r",stdin);	//freopen("out.txt","w",stdout);	int n;	while (~scanf("%d", &n))	{		map
     
       mp;		for (int i = 0; i < n; i++)		{			scanf("%d", a + i);			mp[a[i]]++;		}		int num = 0;		for (int i = 0; i < n; i++)		{			if (mp[a[i]])	//加进去与前面有的筛一下就好			{				mp[a[i]]--;				ans[num++] = a[i];				for (int k = num - 2; k >= 0; k--)					if (mp[ans[num - 1] + ans[k]])						mp[ans[num - 1] + ans[k]]--;			}		}		printf("%d\n", num);		for (int i = 0; i < num; i++)		{			cout << ans[i];			if (i != num - 1) printf(" ");		}		puts("");	}	return 0;}