Determine+the+difference+in+potential+between+two+points+that+are+distances+Ra,+and+Rb+from+a+very+l

The electric potential difference between two points that are distances Ra and Rb from a very long straight wire carrying a uniform charge per unit length λ can be calculated using the formula:

V = λ/2πε₀ ln(Rb/Ra)

where ε₀ is the permittivity of free space.

This formula is derived from the electric field equation for a long straight wire, which is given by:

E = λ/2πε₀r

where r is the distance from the wire.

To find the potential difference between two points, we integrate the electric field equation from one point to the other, which gives us:

Vb - Va = -∫E·ds = -∫λ/2πε₀r·dr

Integrating this expression gives us:

Vb - Va = λ/2πε₀ ln(Rb/Ra)

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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.