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(1) * 4;



using namespace std;

long long base[] = {1,1,3,9,27,81,243,729,2187,6561,19683,59049,177147,531441,1594323,4782969,14348907,43046721,129140163,387420489,1162261467,3486784401};

int a[100];

int tt[110];

int main()

{

    //freopen("input.in","r",stdin);

    //freopen("output.in","w",stdout);

    int n;

    cin >> n;

    while(n--){

        int x;

        cin >> x;

        memset(a,0,sizeof(a));

        memset(tt,0,sizeof(tt));

        int tmp = x;

        while(tmp){

            a[++a[0] ] = tmp % 3;

            tmp /= 3;

        }

        for(int i = 1;i <= a[0];i++){

            if(a[i] == 2){

                a[i+1]++;

                a[0] = max(i+1,a[0]);

                tt[ ++tt[0] ] = i;

                a[i] = 0;

            }

            if(a[i] == 3){

                a[i+1]++;

                a[0] = max(i+1,a[0]);

                a[i] = 0;

            }

        }

        cout << "left pan:";

        for(int i = tt[0];i > 0;i--){

            cout << " " << base[ tt[i] ];

        }

        cout << endl;

        cout << "right pan:";

        for(int i = a[0];i > 0;i--){

            if(!a[i])

                continue;

            cout << " " << base[i];

        }

        cout  << endl;

        if(n)

            cout << endl;

    }

    return 0;

}

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