HDU 6168 Numbers

Numbers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 92    Accepted Submission(s): 44

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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liuyiding

题意：

b数组长度为m(0<=m<=125250)，a数组长度为n，保证n*(n-1)/2恒等于m。ai+aj的值作为b数组中的元素(1<=i<j<=n)，给出b数组，还原出a数组。

思路：

b数组中最小的元素一定是a数组中的，因为b数组中最小的元素是不可能通过中原始的a数组加和得到的，把它们取出来放到a数组中，之后让这些元素和a数组中的元素两两之间相加和，去除b数组中多余元素，操作完以后再找到次小元素，继续上面的操作，直到b数组中没有元素为止。

我利用的是map映射记录b数组的元素以及它们的个数，因为每个数的值可能会比较大。

示例程序

#include <bits/stdc++.h>using namespace std;map<int,int>mp;//记录b数组每一个元素编号所对应的值map<int,int>id;//记录b数组的元素编号int v[125255],a[125255],b[125255];void update(int x,int top){    int i,t;    for(i=0;top>i;i++)    {        t=x+a[i];        if(mp[t]!=0)//如果该元素在b数组中，数量减1        {            v[mp[t]]--;        }    }}int main(){    int m,i,num,top,x;    while(scanf("%d",&m)!=EOF)    {        mp.clear();        id.clear();        num=1;        top=0;        memset(v,0,sizeof(v));        for(i=1;m>=i;i++)        {            scanf("%d",&b[i]);        }        sort(b,b+m);//对b数组排一下序，便于我们从小到大取出元素        for(i=1;m>=i;i++)        {            if(mp[b[i]]==0)            {                mp[b[i]]=num;//该元素没有出现过，对b数组的元素编号                id[num]=b[i];                num++;            }            v[mp[b[i]]]++;//数量加1        }        for(i=1;num>i;i++)        {            while(v[i]>=1)//如果b数组中有该元素，移动到a数组            {                v[i]--;                x=id[i];                update(x,top);//除去多余元素操作                a[top]=x;                top++;            }        }        printf("%d\n",top);        for(i=0;top>i;i++)        {            if(i!=0)            {                printf(" ");            }            printf("%d",a[i]);        }        printf("\n");    }    return 0;}