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 ;

}