Many staff of are living in a place called MZone, far from their office( 4.5 km ). Due to the bad traffic, many staff choose to ride a bike. We may assume that all the people except "Weiwei" ride from home to office at a fixed speed. Weiwei is a people with a different riding habit – he always tries to follow another rider to avoid riding alone. When Weiwei gets to the gate of MZone, he will look for someone who is setting off to the Office. If he finds someone, he will follow that rider, or if not, he will wait for someone to follow. On the way from his home to office, at any time if a faster student surpassed Weiwei, he will leave the rider he is following and speed up to follow the faster one. We assume the time that Weiwei gets to the gate of MZone is zero. Given the set off time and speed of the other people, your task is to give the time when Weiwei arrives at his office.

To solve this problem, we can use the concept of relative speed. Let's consider the scenario step by step:

1. First, we need to find the person who sets off the earliest among all the riders. Let's call this person "earliestRider". The time when Weiwei arrives at his office will be equal to the set off time of "earliestRider" plus the time it takes for Weiwei to catch up with "earliestRider".

2. Next, we calculate the relative speed between Weiwei and "earliestRider". If Weiwei is faster than "earliestRider", he will catch up with them before they reach the office. Otherwise, Weiwei will need to wait for someone faster to catch up to him.

3. Once Weiwei catches up with "earliestRider", we update the set off time of "earliestRider" to the time when Weiwei catches up with them.

4. We repeat steps 1-3 until Weiwei reaches his office.

Here's a sample C++ code that implements this logic:

#include <iostream>
#include <vector>

struct Rider {
    int setOffTime;
    int speed;
};

int main() {
    int n; // number of riders (excluding Weiwei)
    std::cout << "请输入骑行人数（不包括Weiwei）：";
    std::cin >> n;

    std::vector<Rider> riders(n);
    std::cout << "请依次输入每位骑行人的出发时间和速度：" << std::endl;
    for (int i = 0; i < n; i++) {
        std::cin >> riders[i].setOffTime >> riders[i].speed;
    }

    int weiweiSpeed;
    std::cout << "请输入Weiwei的速度：";
    std::cin >> weiweiSpeed;

    int weiweiArrivalTime = 0;

    while (true) {
        int earliestRiderIndex = -1;
        int earliestRiderTime = INT_MAX;

        // Find the earliest rider
        for (int i = 0; i < n; i++) {
            if (riders[i].setOffTime < earliestRiderTime) {
                earliestRiderTime = riders[i].setOffTime;
                earliestRiderIndex = i;
            }
        }

        // Calculate the time Weiwei takes to catch up with the earliest rider
        double timeToCatchUp = static_cast<double>(4.5) / (weiweiSpeed - riders[earliestRiderIndex].speed);

        if (earliestRiderTime + timeToCatchUp <= 60) {
            weiweiArrivalTime = earliestRiderTime + timeToCatchUp;
            riders[earliestRiderIndex].setOffTime += timeToCatchUp;
        } else {
            break; // Weiwei arrives at or after 60 minutes, stop the loop
        }
    }

    std::cout << "Weiwei到达办公室的时间为：" << weiweiArrivalTime << "分钟" << std::endl;

    return 0;
}

In this code, we first input the number of riders (excluding Weiwei) and their set off times and speeds. Then we input Weiwei's speed. The program iterates until Weiwei arrives at or after 60 minutes, finding the earliest rider, calculating the time to catch up with them, and updating their set off time accordingly. Finally, it outputs Weiwei's arrival time at the office.

Please note that this is a simplified implementation and doesn't handle all possible edge cases. You can modify and improve it based on your specific requirements. Let me know if you have any further questions!