HDU 6168 Numbers

Numbers

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 289 Accepted Submission(s): 148

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

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题意：
给一个序列a，对应地有序列b为a的第i,j项的和。现在把a和b混在一起，总的长度为m，求原来的序列a。

思路：
题目中已经给了m=n(n-1)/2，显然 n=2m−−−√ 。然后对这个序列进行排序，最小的两项就是a1,a2。删掉a1,a2,a1+a2，最小的就是a3。删掉a3,a3+a1,a3+a2，最小的就是a4。以此类推。
一开始我的思路是对了，但是拿数组写的，然后TLE了。后来用了vector也TLE，网上说vector的push_back和erase复杂度有点高，果然木有错。改用multiset之后就A了。

代码：

#include<stdio.h>#include<math.h>#include<iostream>#include<algorithm>#include<set>#define ll long long#define maxn 125255using namespace std;multiset<ll> q;multiset<ll> a;ll m,n;int main(){    while(~scanf("%lld",&m))    {        n=sqrt(2*m);        ll i,j;        for(i=0;i<m;i++)        {            ll x;            scanf("%lld",&x);            q.insert(x);        }        printf("%lld\n",n);        a.insert(*(q.begin()));        q.erase(q.begin());        a.insert(*(q.begin()));        q.erase(q.begin());        ll step=1;        multiset<ll>::iterator itor;        while(a.size()<n)        {            for(itor=a.begin();itor!=(--a.end());itor++)            {                q.erase(q.find(*itor+*(--a.end())));            }            a.insert(*(q.begin()));            q.erase(q.begin());        }        for(itor=a.begin();itor!=a.end();itor++)        {            printf("%lld",*itor);            if(itor!=(--a.end())) printf(" ");        }        printf("\n");        q.clear();        a.clear();    }}

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P.S. 我仿佛已经看见公式灵魂漂移的样子了，过段时间自己建站吧QAQ