HDU 6168 Numbers(排序)

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

62 2 2 4 4 4211 2 3 3 4 4 5 5 5 6 6 6 7 7 7 8 8 9 9 10 11

 

Sample Output

32 2 261 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。

#include <iostream>#include <string.h>#include <vector>#include <algorithm>#include <map>#include <stdio.h>using namespace std;#define  LL long longconst 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");    }}