4月 04, 2009

【解題】Combination Lock

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ACM Volume CV 10550 - Combination Lock


The Problem

Now that you're back to school for another term, you need to remember how to work the combination lock on your locker. A common design is that of the Master Brand, shown at right. The lock has a dial with 40 calibration marks numbered 0 to 39. A combination consists of 3 of these numbers; for example: 15-25-8. To open the lock, the following steps are taken:
  • turn the dial clockwise 2 full turns
  • stop at the first number of the combination
  • turn the dial counter-clockwise 1 full turn
  • continue turning counter-clockwise until the 2nd number is reached
  • turn the dial clockwise again until the 3rd number is reached
  • pull the shank and the lock will open.
Given the initial position of the dial and the combination for the lock, how many degrees is the dial rotated in total (clockwise plus counter-clockwise) in opening the lock?


Input

Input consists of several test cases. For each case there is a line of input containing 4 numbers between 0 and 39. The first number is the position of the dial. The next three numbers are the combination. Consecutive numbers in the combination will be distinct. A line containing 0 0 0 0 follows the last case.


Output

For each case, print a line with a single integer: the number of degrees that the dial must be turned to open the lock.


Sample Input

0 30 0 30
5 35 5 35
0 20 0 20
7 27 7 27
0 10 0 10
9 19 9 19
0 0 0 0


Sample Output

1350
1350
1620
1620
1890
1890


解題思考

  這題看起來好像有點複雜,不過找出公式之後問題便可以迎刃而解。(不過感覺很怪異,似乎是題目把順時針跟逆時針寫反了?)


  讓我們先分別求出題目五個步驟的角度值。

  首先,假設每一筆輸入資料的四個號碼分別為 d0、d1、d2、d3。且由於轉盤共有 40 個刻度,所以一個刻度的角度值恰好為 9。
  • 第一步,我們順時針旋轉兩圈。所以旋轉的角度值為:720 度。
  • 第二步,我們由起始數字 d0 順時針旋轉到第一個號碼 d1。所以若是 d0 > d1,旋轉的角度值為:(d0 - d1) × 9 度;若是 d0 < d1,旋轉的角度值為:(d0 - d1 + 40) × 9 度。
  • 第三步,我們逆時針旋轉一圈。所以旋轉的角度值為:360 度。
  • 第四步,我們由第一個數字 d1 逆時針旋轉到第二個號碼 d2。所以若是 d2 > d1,旋轉的角度值為:(d2 - d1) × 9 度;若是 d2 < d1,旋轉的角度值為:(d2 - d1 + 40) × 9 度。
  • 第五步,我們從第二個號碼 d2 順時針旋轉到第三個號碼 d3。所以若是 d2 > d3,旋轉的角度值為:(d2 - d3) × 9 度;若是 d2 < d3,旋轉的角度值為:(d2 - d3 + 40) × 9 度。
  利用程式將以上這些結果相加,就是答案了。


參考解答(C++)

#include <iostream>

using namespace std;

int main(void)
{
    while (1)
    {
        int d[4];
        cin >> d[0] >> d[1] >> d[2] >> d[3];

        if (d[0] == 0 && d[1] == 0 && d[2] == 0 && d[3] == 0)
        {
            break;
        }

        // 計算角度
        int degree = 120;
        degree += (d[0] - d[1] + 40) % 40;
        degree += (d[2] - d[1] + 40) % 40;
        degree += (d[2] - d[3] + 40) % 40;

        cout << degree * 9 << endl;
    }

#ifndef ONLINE_JUDGE
    system("pause");
#endif

    return 0;
}

4 回覆:

kgame 智涵 提到...

轉盤式密碼鎖確實是順逆順321
題目沒寫錯阿

小參 提到...

不, 你拿 Sample Input 的數值手算看看
題目的順逆時針與 Sample Output 的結果不合

kgame 智涵 提到...

0 30 0 30
0順時針到30有10刻度
30逆時針到0共10刻度
0順時針到30有10刻度
共30 * 9 + 1080 = 1350度
沒搞錯阿@@

小參 提到...

是我搞錯了, 不好意思 囧>

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