POJ 1702 Eva's Balance_eva has a balance with 20 poises. the weights of t-优快云博客

本文链接：https://blog.youkuaiyun.com/bobten2008/article/details/4818097

本文介绍了一种使用具有特定权重的砝码测量不同物体重量的方法。通过将待测物体的重量转换为三进制，并利用特定算法放置砝码，可以准确测量出1到(3^20-1)/2范围内的任意整数重量。

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Eva's Balance

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2889		Accepted: 1424

Description

Eva has a balance with 20 poises. The weights of the poises are 1, 3, 9, 27,...,3^19. Eva asserts that she has a way to measure any object whose weight is an integer from 1 to (3^20-1)/2. Assuming that Eva has placed an object with the weight in this range on the left tray of the balance, your task is to place the proper poises on the proper trays so as to weigh the object out.

Input

The first line is an integer T (1 <= T <= 20), which shows the number of the test cases. Each of the following T lines contains an integer W (1 <= W <= (3^20-1)/2), expressing the weight of an object.

Output

For each test case print a line, showing the weights of the poises on the left tray and the right tray. Use a space to separate the left tray and the right tray. The poises on the same tray are arranged in the increasing order, and a comma separates every two of them. If there is no poise on the tray, output "empty".

Sample Input

Sample Output

empty 9
1,3 9
1,9 3,27

Source

POJ Monthly--2004.07.18

/* 先将输入数num转换成三进制tinary, 然后从低位往高位处理三进制, tinary[i]其值有四种可能: 1)0, 跳过 2)1, 将3 ^ i加入右序列 2)2, 将3 ^ i加入左序列并将tinary[i + 1]++ 3)3, 将tinary[i + 1]++; 为什么能表示的最大数是(3^20-1)/2呢, 由于一共只有20位,所以最大只能表示全1的数,既20个1, 已经是最大的了, 往任意位置上加1都导致需要超过20位来表示,运用等比数列求和所以能表示的最大数是(3^20-1)/2 */ #include <iostream> #define MAX_N 20 int expv[MAX_N + 1]; int tinary[MAX_N + 1]; int left[MAX_N + 1]; int right[MAX_N + 1]; int lnum, rnum; int getTinary(int n) { int quotient = n, len = 0; while(quotient != 0) { tinary[len++] = quotient % 3; quotient /= 3; } return len; } void getPos(int n) { lnum = rnum = 0; memset(tinary, 0, sizeof(tinary)); int p, len = getTinary(n); for(p = 0; p < len; p++) { if(tinary[p] == 1) right[rnum++] = expv[p]; else if(tinary[p] == 2) { left[lnum++] = expv[p]; tinary[p + 1]++; } else if(tinary[p] == 3) tinary[p + 1]++; } if(tinary[len] == 1) right[rnum++] = expv[len]; if(lnum == 0) printf("empty"); else { for(p = 0; p < lnum; p++) { if(p != 0) printf(","); printf("%d", left[p]); } } printf(" "); if(rnum == 0) printf("empty"); else { for(p = 0; p < rnum; p++) { if(p != 0) printf(","); printf("%d", right[p]); } } printf("/n"); } int main() { int caseN, i; expv[0] = 1; for(i = 1; i < MAX_N; i++) expv[i] = expv[i - 1] * 3; scanf("%d", &caseN); int num; while(caseN--) { scanf("%d", &num); getPos(num); } return 0; }