Problem B: Ternarian Weights

大致题意：使用三进制砝码采取相应的措施衡量出给定的数字
主要思路：三进制，如果 大于 2 向前进位，之前一直没写好放弃了，这次终于写好了……

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <map>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <cctype>

const double Pi = atan() * ;

using namespace std;
long long base[] = {,,,,,,,,,,,,,,,,,,,,,};
int a[];
int tt[];
int main()
{
    //freopen("input.in","r",stdin);
    //freopen("output.in","w",stdout);
    int n;
    cin >> n;
    while(n--){
        int x;
        cin >> x;
        memset(a,,sizeof(a));
        memset(tt,,sizeof(tt));
        int tmp = x;
        while(tmp){
            a[++a[] ] = tmp % ;
            tmp /= ;
        }
        for(int i = ;i <= a[];i++){
            if(a[i] == ){
                a[i+]++;
                a[] = max(i+,a[]);
                tt[ ++tt[] ] = i;
                a[i] = ;
            }
            if(a[i] == ){
                a[i+]++;
                a[] = max(i+,a[]);
                a[i] = ;
            }
        }
        cout << "left pan:";
        for(int i = tt[];i > ;i--){
            cout << " " << base[ tt[i] ];
        }
        cout << endl;
        cout << "right pan:";
        for(int i = a[];i > ;i--){
            if(!a[i])
                continue;
            cout << " " << base[i];
        }
        cout  << endl;
        if(n)
            cout << endl;
    }
    return ;
}