一个关于C++的问题

Question

首先给你一个数字T，标示T组数据。
接下来T行，每行两个数字x和y，标示每次计时的开始时间和结束时间。
对于每次计时，你需要计算开始时间到结束时间在钟面上时针和分针会碰面多少次，并输出
样例输入
2
14 17
5 9
样例输出
3
4
帮我用C++做一下，谢！

Answer 1

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To solve this problem, we can use the following approach:

1. Initialize a counter variable count to 0 and two variables start_time and end_time representing the start time of the first time point (i.e., start_time = 0, end_time = T) and the end time of the last time point (i.e., end_time = T - 1).

2. Iterate through each row in the input data:

   • Read two numbers from the x and y columns.
   • Add x and y as the coordinates for the current time point by subtracting the minimum value from start_time and maximum value from end_time.
   • Increment the count variable by 1.

3. Calculate the number of times the left-hand pointer (i.e., minutes) is less than or equal to the right-hand pointer (i.e., seconds) using the modulus operator %. Let's call this condition left_count.

4. For each time point:

   • If left_count is greater than 0, add the time difference between the current time and the previous time point to the end_time variable.
   • Subtract start_time from end_time to get the actual elapsed time in minutes.

5. Generate a list of all possible pairs of minute values where the hour and minute are distinct (since multiple minutes can overlap within an hour). This list will be represented as a set in C++.

6. Calculate the number of overlapping pairs by counting the number of unique pair combinations that satisfy both conditions: num_overlapping_pairs = count * (left_count + 1) / 2. The (left_count + 1) factor accounts for the fact that we need to consider the extra minute at the end of the day.

7. Print the result by outputting the value of num_overlapping_pairs.

Here's the implementation in C++:

#include <iostream>
#include <vector>
#include <set>

int main() {
    int T;
    std::cin >> T;

    // Initialize counters
    int count = 0;
    int start_time = T;
    int end_time = T - 1;

    // Iterate through rows
    for (int i = 0; i < T; ++i) {
        std::cin >> x[i] >> y[i];

        // Calculate coordinates for the current time point
        start_time -= min(x[i], end_time);
        end_time -= max(x[i], end_time);

        // Increment the count
        count++;

        // Calculate the left hand pointer (minutes)
        if (start_time % 60 <= end_time % 60) {
            left_count = end_time % 60 - start_time % 60;
        }

        // Determine hours and minutes for the time point
        int hour = (start_time / 60) % 24;
        int minute = (start_time / 60) % 60;

        // Generate a list of pairs with distinct hour and minute values
        std::vector<std::pair<int, int>> time_pairs;
        while (left_count > 0) {
            if ((minute - 1) % 5 == 0 && hour == 12) { // Zero-based hour positions
                time_pairs.push_back(std::make_pair(minute + 1, hour));
                break;
            }
            if ((minute - 1) % 5 == 0 || minute % 5 != 0) { // Non-zero-based minute positions
                time_pairs.push_back(std::make_pair(minute, hour));
            } else {
                time_pairs.push_back(std::make_pair(minute, hour - 1));
                break;
            }
            left_count -= 1;
        }

        // Calculate the number of overlapping pairs
        num_overlapping_pairs = count * (left_count + 1) / 2;

        // Output the result
        std::cout << num_overlapping_pairs << std::endl;
    }

    return 0;
}

Please note that this code assumes that x and y columns represent hours (0-23) and minutes (0-59), respectively, and that the right-hand pointer represents seconds. You may need to modify the code based on your specific requirements or data format if it differs. Additionally, the code only considers one hour per row, so if you have multiple hours for a given time point, you'll need to modify the logic accordingly. Finally, you should compile and run the program with the provided input data to verify the solution.