Hdu 6168 Numbers【思维+暴力】

Numbers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 71    Accepted Submission(s): 34

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

题目大意：

如果我们现在有一个数组A【】，那么我们每一次拿出两个数A【i】和A【j】，使得B【cnt++】=A【i】+A【j】，主人公不小心把两个数组并到了一起，并且还是从小到大保证递增的，问我们数组A【】的信息。输出其大小和元素。

思路：

考虑到A【】数组的最大长度是500.那么我们暴力去搞就行。

最小的数肯定属于A【】，理所当然倒数第二小的数也肯定属于A【】，那么我们就可以根据这两个数确定一个B【】肯定有的，那么我们将全部序列中去掉这几个数之后，肯定接下来那个最小的数也属于A【】；依次类推，搞搞就行了。

时间复杂度O(n+500^2)；

Ac代码：

#include<stdio.h>#include<string.h>#include<map>using namespace std;int a[125255];int output[125255];int main(){    int n;    while(~scanf("%d",&n))    {        int ans=0;        for(int i=1;i<=n;i++)        {            if(i+(i)*(i-1)/2==n)            {                ans=i;            }        }        map<int,int>s;        for(int i=1;i<=n;i++)scanf("%d",&a[i]);        int cnt=0;        for(int i=1;i<=n;i++)        {            if(s[a[i]]>0)            {                s[a[i]]--;            }            else            {                for(int j=0;j<cnt;j++)                {                    s[a[i]+output[j]]++;                }                output[cnt++]=a[i];            }        }        printf("%d\n",cnt);        for(int i=0;i<cnt;i++)        {            if(i>0)printf(" ");            printf("%d",output[i]);        }        printf("\n");    }}