hdu 6168 Numbers（多校联赛）

Numbers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2015    Accepted Submission(s): 710

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
题意：给出a和b的混合数列，其中a数列长度为n，b数列长度为n*(n-1)/2;b数列中的项为ai+aj的值，
现给出a,b的混合数列，求a数列。
#include<iostream>#include<bits/stdc++.h>using namespace std;const int N=1e5+10;int a[N];int main(){    int i,j,n,m;    while(scanf("%d",&m)!=EOF)    {        priority_queue<int,vector<int>,greater<int> >q1;        priority_queue<int,vector<int>,greater<int> >q2;        for(int i=1;i<=m;i++)        {            if(i*(i+1)==m*2)            {                n=i;                break;            }        }        printf("%d\n",n);        int c;        for(int i=0;i<m;i++)        {            scanf("%d",&c);            q1.push(c);        }        int cnt=1;        a[0]=q1.top(),q1.pop();        a[1]=q1.top(),q1.pop();        q2.push(a[0]+a[1]);        while(!q1.empty())        {            while(1)            {                if(q1.top()==q2.top())                {                    q1.pop();                    q2.pop();                }                if(q2.empty()==true||q2.top()!=q1.top());                {                    cnt++;                    a[cnt]=q1.top();                    q1.pop();                    break;                }            }            if(cnt==n-1)            {                break;            }            for(int i=0;i<cnt;i++)            {                q2.push(a[cnt]+a[i]);            }        }        for(int i=0;i<n;i++)        {            if(i==0)                printf("%d",a[i]);            else                printf(" %d",a[i]);        }        printf("\n");    }    return 0;}