HDU—6168（Numbers）

Numbers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 907    Accepted Submission(s): 479

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

题意分析：有n个数字（a[1]、a[2]、a[3]···）将其中任意两个数相加得到序列b，b[1] = a[1]+a[2]、b[2] = a[1]+a[3]、b[3] = a[1]+a[4]···b[n*(n-1)/2] = a[n-1]+a[n]，现将a序列和b序列打乱顺序混合在一起，得到c序列，现在已知c序列，求a序列；

n](n−1

解题分析：由于b序列是由a序列元素求和得到的，故序列c排序后前两个元素c[0]、c[1]即为序列a的前两个元素，同时，既然b序列是由a序列求和得到的，统计利用已有的a[0]、a[1]元素，便可求出b[1]，那么c序列中可删除其中同等数量的b[1]元素，剩下的b[1]则表明是属于序列a的元素，将其送入a序列中。利用同样方法利用已有a元素值，求得b元素值个数。在c中比较，剩余则存入a中。

ＡＣ　ｃｏｄｅ：

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
using namespace std;
int main (void){
    int length;
    while(scanf("%d",&length) != EOF){
        int i,j;
        //利用map统计每个元素的个数 
        map<int,int>count;
        int number = 0;
        vector<int>c;
        vector<int>a;
        if(length == 0){
            printf("%d\n",0);
            continue;
        } 
        for(i=0;i<length;i++){
            scanf("%d",&number);
            c.push_back(number) ;
            //统计元素个数 ，最少出现次数为一 
            if(count[number] == 0) 
                count[number] = 1;
            else count[number]++;      
        }
        for(i=0;i<length;i++){
        	//为 0 则说明其不是a序列元素 
            if(count[c[i]] == 0 ) continue;
            for(j=0;j<a.size();j++)
            {
            	//a序列自身能求出的和在C中自减，剩下的和为a序列的和 
                count[ a[j] + c[i] ]--;
            }
            a.push_back(c[i]); 
            count[c[i]]--;
        }
        printf("%d\n",a.size());
        printf("%d",a[0]);
        for(i=1;i<a.size();i++)
           printf(" %d",a[i]); 
        printf("\n");
    }
    return 0;
}