hdu6168

Numbers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 81    Accepted Submission(s): 38

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：

有n个数a，可以两两组合成n（n-1）／2个数b。把它们混成一个数组，让你求出原n个数a。

POINT：

排个序，第一小和第二小肯定都是a数组里的。开一个map记录当前可以确定的b数组里的数。

一开始map里只有a[1]+a[2]。

从3-n开始找，若map里有，则是b数组，更新map。若没有，即是a数组里的，并且把和之前a数组里的数组成的数b更新在map里。

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print ?

1. #include <iostream>  
2. #include <string.h>  
3. #include <vector>  
4. #include <algorithm>  
5. #include <map>  
6. #include <stdio.h>  
7. using namespace std;  
8. #define  LL long long  
9. const int N = 125250+8;  
10. map<int,int> m;  
11. vector<int> a;  
12. int z[N];  
13. int main()  
14. {  
15.     int n;  
16.     while(~scanf("%d",&n))  
17.     {  
18.         m.clear();  
19.         a.clear();  
20.         for(int i=1;i<=n;i++)  
21.         {  
22.             scanf("%d",&z[i]);  
23.         }  
24.         sort(z+1,z+1+n);  
25.         a.push_back(z[1]);  
26.         a.push_back(z[2]);  
27.         m[z[1]+z[2]]++;  
28.         for(int i=3;i<=n;i++)  
29.         {  
30.             if(m[z[i]]>0)  
31.             {  
32.                 m[z[i]]--;  
33.                 continue;  
34.             }  
35.             else  
36.             {  
37.                 a.push_back(z[i]);  
38.                 for(int j=0;j<a.size()-1;j++)  
39.                 {  
40.                     m[a[j]+z[i]]++;  
41.                 }  
42.             }  
43.         }  
44.         sort(a.begin(),a.end());  
45.         printf("%d\n",a.size());  
46.         for(int i=0;i<a.size();i++)  
47.         {  
48.             if(i) printf(" ");  
49.             printf("%d",a[i]);  
50.         }  
51.         printf("\n");  
52.     }  
53. }  

#include <iostream>
#include <string.h>
#include <vector>
#include <algorithm>
#include <map>
#include <stdio.h>
using namespace std;
#define  LL long long
const int N = 125250+8;
map<int,int> m;
vector<int> a;
int z[N];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        m.clear();
        a.clear();
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&z[i]);
        }
        sort(z+1,z+1+n);
        a.push_back(z[1]);
        a.push_back(z[2]);
        m[z[1]+z[2]]++;
        for(int i=3;i<=n;i++)
        {
            if(m[z[i]]>0)
            {
                m[z[i]]--;
                continue;
            }
            else
            {
                a.push_back(z[i]);
                for(int j=0;j<a.size()-1;j++)
                {
                    m[a[j]+z[i]]++;
                }
            }
        }
        sort(a.begin(),a.end());
        printf("%d\n",a.size());
        for(int i=0;i<a.size();i++)
        {
            if(i) printf(" ");
            printf("%d",a[i]);
        }
        printf("\n");
    }
}