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

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

 

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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[N], b[N];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;}