The material properties never play a role when one tries to determine the support reactions of a structure under loading.

This statement is incorrect. The material properties of a structure play a crucial role in determining its support reactions under loading. The support reactions of a structure depend on factors such as the type of material used, its strength, stiffness, and other properties. For example, a structure made of steel will have different support reactions compared to a similar structure made of wood or concrete, due to differences in their material properties. Therefore, it is important to consider the material properties when analyzing the support reactions of a structure.

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Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 ≤ L ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distance Di from the start (0 < Di < L). To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river. Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M rocks (0 ≤ M ≤ N). FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks. Input Line 1: Three space-separated integers: L, N, and M Lines 2..N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position. Output Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks Sample Inputcopy Outputcopy 25 5 2 2 14 11 21 17 4 Hint Before removing any rocks, the shortest jump was a jump of 2 from 0 (the start) to 2. After removing the rocks at 2 and 14, the shortest required jump is a jump of 4 (from 17 to 21 or from 21 to 25).

To determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks, we can follow these steps:

1. Read the input values for L, N, and M.
2. Read the distances of the N rocks from the starting rock and store them in a list.
3. Sort the list of distances in ascending order.
4. Calculate the initial shortest jump as the distance from the starting rock to the first rock.
5. Initialize a variable max_jump to store the maximum shortest jump.
6. Iterate over each possible combination of removing M rocks from the list of distances. We can use a nested loop to generate all combinations.
- For each combination, calculate the shortest jump after removing the selected rocks.
- Update max_jump if the current shortest jump is greater than max_jump.
7. Print the value of max_jump as the maximum shortest distance a cow has to jump after removing M rocks.

Here's an example implementation in Python:

from itertools import combinations

L, N, M = map(int, input().split())

rocks = []
for _ in range(N):
    rocks.append(int(input()))

rocks.sort()

initial_jump = rocks[0]
max_jump = 0

for remove_rocks in combinations(range(1, N + 1), M):
    jumps = [rocks[remove_rocks[i]] - rocks[remove_rocks[i - 1] - 1] for i in range(1, M)]
    jumps.append(L - rocks[remove_rocks[M - 1] - 1])
    shortest_jump = min(jumps)
    max_jump = max(max_jump, shortest_jump)

print(max_jump)

In the example input provided, the output would be `4`, which represents the maximum shortest distance a cow has to jump after removing 2 rocks.

Note: This solution uses brute force to iterate over all possible combinations of removing M rocks. The time complexity is O(N choose M), which can be large for large values of N and M.

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!