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): 13    Accepted Submission(s): 9

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

比赛也快结束了 感觉自己越来越菜了 就做出来一道题 写一下今天唯一的成果吧 只做出来了1008 

题意：给你一个队列 这个队列是由数列a和b混合起来的 a数列长度为n b数列长度为n（n-1）/2期中b数列为ai+aj i和j在1到n的范围之内 让你根据给出的数列求出a数列来 

思路：刚开始做的时候 暴力一发然后tle了 人后听别的队用优先队列做的 于是也改成了优先队列  因为给你的是一个混合的总的数列 所以要根据a数列 把b数列踢出 剩下的就都是a数列了 而且这一个过程要一边踢一边存 这样才不会对后续的过程产生影响 但是队列又不能从中间删除元素 所以用一个辅助的队列来存应该被删除的数

过程：把序列存到优先队列去 排序后数列的前两个肯定是a数列的前两个元素 弹出这两元素放到集合a里去 然后把这两个数生成的b放到辅助队列中 让辅助队列顶元素去和主队列顶元素对比 如果相同 就把这个元素都弹出去 如果不相同就把主队列的栈顶元素弹出来加入a集合 与集合a中其他的元素生成新的b数列的元素 并压入到辅助数列中去 然后重复就ok了

ac代码：

#include<cstdio>#include<iostream>#include<queue>#include<vector>#include<cstring>#include<cmath>using namespace std;const int N=1e5+10;int arr[N];int main(){    int m;    while(scanf("%d",&m)!=EOF)    {        priority_queue<int,vector<int>,greater<int> >q1;        priority_queue<int,vector<int>,greater<int> >q2;        int n=(-1+(int)sqrt((double)(1+8*m)))/2;        printf("%d\n",n);        int a;        for(int i=0;i<m;i++)        {            scanf("%d",&a);            q1.push(a);        }        int cnt=1;        arr[0]=q1.top(),q1.pop();        arr[1]=q1.top(),q1.pop();        q2.push(arr[0]+arr[1]);        while(!q1.empty())        {            while(1)            {                if(q1.top()==q2.top())               {                q1.pop();                q2.pop();               }               if(q2.empty()==true||q2.top()!=q1.top())                {                   cnt++;                   arr[cnt]=q1.top();                   q1.pop();                   break;                }            }            if(cnt==n-1)                break;            for(int i=0;i<cnt;i++)            {                q2.push(arr[cnt]+arr[i]);            }        }        for(int i=0;i<n;i++)        {            if(i==0) printf("%d",arr[i]);            else printf(" %d",arr[i]);        }        printf("\n");    }    return 0;}