D. Divide by three, multiply by two

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本文介绍了一个关于整数操作的算法题目实现，通过遍历输入的整数序列，寻找可以被三整除且乘以二或能被二整除且除以二的数对，并逆向追踪这些操作，最终输出原始起点的整数。

D. Divide by three, multiply by two

//map，long long int
#include<bits/stdc++.h>
using namespace std;
vector<long long int>a;
map<long long int , int>pre;
map<long long int, long long int>prea;
map<long long int , long long int>preb;
int main()
{
    int n;
    scanf("%d", &n);
    for(int i=0;i<n;i++)
    {
        long long int x;
        scanf("%lld", &x);
        a.push_back(x);
        pre[x]=1;
    }
    sort(a.begin(), a.end());
    long long int point;
    for(int i=0;i<n;i++)
    {
       long long int x = a[i]*3;
       long long int y = 0;
       if(a[i]%2==0)
       {
           y = a[i]/2;
       }
     //  cout<<a[i]<<"@@@@@@@"<<x<<"****"<<y<<"^^^^^^"<<pre[x]<<"^^^^"<<pre[y]<<endl;
       if(pre[x])
           {
               prea[x] = a[i];
               preb[a[i]] = x;
           }
       else if(pre[y])
       {
           prea[y] = a[i];
           preb[a[i]] = y;
       }
       else
        point = a[i];
    }
    for(int i=0;i<n;i++)
    {
       // printf("**%lld %lld**\n", prea[a[i]],  preb[a[i]]);
        printf("%lld%c", point ,i==n-1?'\n':' ');
        point = prea[point];
    }
    return 0;
}