UVa 12468 – Zapping
I'm a huge TV fan, but I don't like sticking to one channel. I often switch between different channels.
My dog chewed up my remote control, and now the number buttons don't work. I'm left with only two buttons to change channels: one for going up (△ button) and one for going down (▽ button). It's really annoying because if I want to switch from channel 3 to channel 9, I have to press the △ button 6 times.
我的電視有 100 個頻道，號碼為 0 到 99。它們是循環的，也就是從 99 台再按一下 △ 就會回到第 0 台。同理，從第 0 台按一下 ▽ 就會回到 99 台。
Can you help me write a program that, given the current channel I'm watching and the channel I want to switch to, tells me the minimum number of button presses needed?
Sample Inputs/Outputs
| Sample Input(s) | Sample Output(s) |
|---|---|
| The input consists of multiple test cases, terminated by EOF (End of File). Up to 200 test cases may be present. Each test case consists of a single line containing two integers, a and b. a represents the current channel, and b represents the channel to switch to. Both a and b are non-negative integers not exceeding 99. The last line consists of two -1 integers and should not be processed. | For each set of data, output an integer on a single line, representing the minimum number of button presses required to switch to the new channel. (Remember, I only have two buttons available: △ and ▽). |
| 3 9 0 99 12 27 -1 -1 | 6 1 15 |
Thought Process
Calculate both the values of "△" and "▽", then compare them to find the smaller value and output that smaller value.
Sample Code－ZeroJudge A518: Zapping
#include <iostream>
using namespace std;
int main() {
int a, b;
while (cin >> a >> b)
{
if (a == -1 && b == -1) break;
else
{
if (a > b) swap(a, b);
int one = b - a;
int two = a - b + 100;
cout << min(one, two) << endl;
}
}
}
//ZeroJudge A518
//Dr. SeanXD