Maximizing Triples Product

Given three integers, you can increment any of them exactly five times. The goal is to maximize their final product. To achieve the maximum product, we should always increment the smallest of the three integers. This minimizes the disparity between the values, which yields the highest product for a fixed sum.

def compute_max_product(x, y, z):
    values = [x, y, z]
    for _ in range(5):
        min_index = values.index(min(values))
        values[min_index] += 1
    return values[0] * values[1] * values[2]

test_count = int(input())
for _ in range(test_count):
    a, b, c = map(int, input().split())
    print(compute_max_product(a, b, c))

Consolidating Fragments

You are given an integer n and a set of fragments that sum up to n. You can perform two operations: split a fragment of size v into v fragments of size 1 (costing v-1 operasions), or merge two fragments of size 1 into a single fragment of size 2 (costing 1 operation). The task is to find the minimum operations to combine all fragments into one piece of size n. Instead of simulating the splits and merges, we can deduce a direct formula. Any fragment of size 1 only needs to be merged. A fragment of size v > 1 requires v-1 splits to break it down into ones, and then v merges to integrate it. This totals 2v - 1 operations for each fragment greater than 1. However, the largest existing fragment does not need to be split, saving 2*max_val - 1 operations. An equivalent simplified formula calculates the total splits as n - max_val and the total merges as k - 1, resulting in 2 * (n - max_val) - (k - 1).

def determine_min_ops(target_size, piece_count, pieces):
    largest_piece = max(pieces)
    splits_needed = target_size - largest_piece
    merges_needed = piece_count - 1
    return 2 * splits_needed - merges_needed

iterations = int(input())
for _ in range(iterations):
    n, k = map(int, input().split())
    arr = list(map(int, input().split()))
    print(determine_min_ops(n, k, arr))

Constructing Optimal Permutations

Construct a permutation of length n where two functions are defined based on thresholds m and k. We want to maximize the cumulative difference between elements evaluated by the first function and those evaluated by the second. To maximize this difference, the elements targeted by the first function should appear as large as possible, while those targeted by the second should appear as small as possible. We can fulfill this by arranging all integers larger than threshold m in descending order, followed by the remaining integers from 1 up to m in ascending order. This guarantees the largest values are attributed to the positive sum and the smallest to the negative sum.

def generate_permutation(size, threshold_m, threshold_k):
    upper_segment = list(range(size, threshold_m, -1))
    lower_segment = list(range(1, threshold_m + 1))
    return upper_segment + lower_segment

cases = int(input())
for _ in range(cases):
    n, m, k = map(int, input().split())
    perm = generate_permutation(n, m, k)
    print(' '.join(map(str, perm)))

River Crossing Simulation

A character must traverse a river from the left bank (position 0) to the right bank (position n+1). The river contains logs (L), water (W), and crocodiles (C). The character can jump up to m meters from a bank or a log, and can swim through water but has a maximum swimming distance of k meters. Crocodiles cannot be jumped on or swam over. We can utilize dynamic programming to track the maximum remaining swimming capacity at each reachable position. Initialize the left bank with k swimming capacity. For each position, if it's a crocodile, it remains unreachable. Otherwise, we check all positions within a m-meter jump radius. If a previous position is a log or the left bank, we can transfer its swimming capacity. Additional, if the previous position was water and had remaining swim capacity, we can swim forward, decrementing the capacity by 1. If the right bank achieves a non-negative capacity, the crossing is possible.

import sys

def can_cross_river(river_len, jump_dist, swim_limit, terrain):
    # dp[pos] tracks max remaining swim capacity upon reaching index pos
    # 0: left bank, 1..river_len: river, river_len+1: right bank
    capacity = [-1] * (river_len + 2)
    capacity[0] = swim_limit
    
    for pos in range(1, river_len + 2):
        if pos <= river_len and terrain[pos - 1] == 'C':
            continue  # Crocodiles block progression
            
        # Evaluate jumps from previous logs or the left bank
        for step in range(1, jump_dist + 1):
            prev_pos = pos - step
            if prev_pos >= 0 and (prev_pos == 0 or (prev_pos <= river_len and terrain[prev_pos - 1] == 'L')):
                if capacity[prev_pos] != -1:
                    capacity[pos] = max(capacity[pos], capacity[prev_pos])
                    
        # Evaluate swimming from a previous water cell
        if pos > 1 and pos <= river_len + 1:
            prev_pos = pos - 1
            if prev_pos <= river_len and terrain[prev_pos - 1] == 'W':
                if capacity[prev_pos] > 0:
                    capacity[pos] = max(capacity[pos], capacity[prev_pos] - 1)
                    
    return "YES" if capacity[river_len + 1] >= 0 else "NO"

def solve():
    input_data = sys.stdin.read().split()
    idx = 0
    t = int(input_data[idx]); idx += 1
    for _ in range(t):
        n = int(input_data[idx]); idx += 1
        m = int(input_data[idx]); idx += 1
        k = int(input_data[idx]); idx += 1
        s = input_data[idx]; idx += 1
        print(can_cross_river(n, m, k, s))

solve()