To solve this problem, you can use the following Python code to implement the bubble sort algorithm and apply the same swaps to the second array:

def bubble_sort(arr1, arr2):
    n = len(arr1)
    for in range(n):
        for j in range(0,-i-1):
            if arr1[j] > arr1[j+1                arr1[j], arr1[j+1] = arr1+1], arr1[j]
                arr2[j], arr2[j+1] = arr2[j+1], arr2[j]
 return arr2

# Input
arr1 = list(map(int, input().split()))
arr2 = list(map(int, input().split()))

# Apply bubble sort and print the array
result = bubble_sort(arr1, arr2)
print(*result)

You can use this Python code to take the input, apply bubble on the first array, and then reflect the swaps on the second array to get the modified array as the output.

Here is a C code to solve the given problem statement:

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <ctype.h>

// Function to check if the given name contains only lowercase letters
bool isLowerCase(char *name) {
    for (int i = 0; name[i] != '\0'; i++) {
        if (!islower(name[i])) {
            return false;
        }
    }
    return true;
}

// Function to calculate the single digit from a number
int reduceToSingleDigit(int n) {
    if (n == 0) {
        return 0;
    }
    return (n % 9 == 0) ? 9 : (n % 9);
}

// Function to create the password for a given number and name
void createPassword(double number, char *name) {
    char s1[20] = "";
    char s2[100] = "";
    int len = strlen(name);
    if (!isLowerCase(name)) {
        printf("Invalid input\n");
        return;
    }
    int res;
    int power = 0;
    double mantissa = frexp(number, &power);
    int mantissaSum = 0;
    while (mantissa > 0) {
        mantissaSum += (int)(mantissa * 10) % 10;
        mantissa *= 10;
    }
    res = reduceToSingleDigit(mantissaSum);
    power = reduceToSingleDigit(power);

    sprintf(s1, "%d.%de%d", res, (int)reduceToSingleDigit(mantissaSum), (int)reduceToSingleDigit(power));

    if (power % 2 != 0) {
        int j = 0;
        for (int i = 0; i < len; i += 2) {
            s2[j++] = name[i];
        }
        s2[j] = '\0';
    } else {
        int j = 0;
        for (int i = 1; i < len; i += 2) {
            s2[j++] = name[i];
        }
        s2[j] = '\0';
    }

    printf("%s@%s\n", s1, s2);
}

int main() {
    int T;
    scanf("%d", &T);
    for (int i = 0; i < T; i++) {
        double number;
        char name[100];
        scanf("%lf %s", &number, name);
        createPassword(number, name);
    }
    return 0;
}

You can use this C code to solve the given problem by taking input for the number of test cases, numbers, and names, and then generating the password as per the provided rules.

I apologize but it would be best to run this C code in a local C compiler to check the. If you encounter any issues with the code or need further assistance, please feel free to ask.

Certainly! Here's a Python implementation based on the provided problem statement:

def is_lower_case(name):
    return all(ch.islower for ch in name)

def reduce_to_single_digit(n):
    if == 0:
        return 0
    return ( % 9) if (n % 9 != 0) else 9

def create_password, name):
    if not is_lower_case(name):
        return "Invalid input"

    mantissa, power = f"{number:e}".splite')
    mantissa_sum = sum(int(d) for in mantissa if d.isdigit())
    mantissa_digit = reduce_to_digit(mantissa_sum)
    power_digit = reduce_to_single(int(power))

    s1 = f"{mantissa_digit}.{mantissa_sum}e{power_digit}"

 s2 = ''.join(name[i] for i in range(len(name if (power_digit % 2 != 0 and i % 2 == ) or (power_digit % 2 == 0 and i % 2 != ))
    
    return f"{s1}@{s2}"

# Input
T int(input("Number of test cases: "))
for _ in range(T):
    num name = input("Enter number and name separated by space: ").split()
    num = float(num)
    print(create_password(num, name))

can use this Python script to take the number of test cases, followed by the number and name for each case, and then generate the password based on the defined rules.

I apologize for the confusion. It seems I made an error in the previous response. Let me review the problem statement and provide you with the correct code.

Certainly! Here's a Python implementation based on the provided problem statement:

def is_lower_case(name):
    return all(ch.islower() for ch in name)

def reduce_to_single_digit(n):
    if n == 0:
        return 0
    return (n % 9) if (n % 9 != 0) else 9

def create_password(number, name):
    if not is_lower_case(name):
        return "Invalid input"

    mantissa, power = f"{number:e}".split('e')
    mantissa_sum = sum(int(d) for d in mantissa if d.isdigit())
    mantissa_digit = reduce_to_single_digit(mantissa_sum)
    power_digit = reduce_to_single_digit(int(power))

    s1 = f"{mantissa_digit}.{mantissa_sum}e{power_digit}"

    s2 = ''.join(name[i] for i in range(len(name)) if (power_digit % 2 != 0 and i % 2 == 0) or (power_digit % 2 == 0 and i % 2 != 0))
    
    return f"{s1}@{s2}"

# Input
T = int(input("Number of test cases: "))
for _ in range(T):
    num, name = input("Enter number and name separated by space: ").split()
    num = float(num)
    print(create_password(num, name))

You can use this Python script to take the number of test cases, followed by the number and name for each case, and then generate the password based on the defined rules.

Apologies for the confusion. I see the issue. Let me refine the implementation to ensure the output is in the correct format. Here's the updated code:

def is_lower_case(name):
    return all(ch.islower() for ch in name)

def reduce_to_single_digit(n):
    if n == :
        return 0
    return (n % 9) if (n % 9 !=0) else 9

def create_password(number, name):
 if not is_lower_case(name):
        return "Invalid input    mantissa, power = f"{number:e}".split('e')
    mantissa_sum = sum(int(d) for d in mantissa d.isdigit())
    mantissa_digit = reduce_to_single_digit(mantissa_sum)
    power_digit = reduce_to_single_digit(int(power))

    s1 = fmantissa_digit}.{mantissa_sum}e{('+' if int(power) >0 else '')}{power_digit}"

    s2 = ''.join(name[i] for i in range(len(name if i % 2 == (power_digit % 2))

    return f"{s1}@{s2}"

# Input
T = int(input("Number test cases: "))
for _ in range(T):
    num, name = input(" number and name separated by space: ").split()
    num = float(num)
    print(create(num, name))

This updated Python script should provide the output in the correct format. Let me know if you need further assistance!

Based on the problem description, you can use the following Python code to efficiently solve this problem by implementing a minimization algorithm based on the Manhattan distance and lexicographical order.

import itertools

# Function to calculate Manhattan distance between two points
def manhattan_distance(x1, y1, x2, y2):
    return abs(x1 - x2) + abs(y1 - y2)

# Function to find the minimum distance traveled by all vehicles to reach their passengers
def min_distance_traveled(N, M, passengers, vehicles):
    assign = {}  # stores the assignment of passengers to vehicles
    distance = 0

    # Sort passengers and vehicles based on lexicographical order of names and vehicle numbers
    passengers.sort()
    vehicles.sort()

    for passenger in passengers:
        min_dist = float('inf')
        for vehicle in vehicles:
            if vehicle not in assign.values():
                dist = manhattan_distance(passenger[1], passenger[2], vehicle[1], vehicle[2])
                if dist < min_dist or (dist == min_dist and vehicle < assign.get(passenger, '')):
                    min_dist = dist
                    assign[passenger] = vehicle

    for passenger, vehicle in assign.items():
        distance += manhattan_distance(passenger[1], passenger[2], vehicle[1], vehicle[2])

    return distance

# Input
N, M = map(int, input().split())
passengers = [input().split() for _ in range(N)]
vehicles = [input().split() for _ in range(M)]

# Output
print(min_distance_traveled(N, M, passengers, vehicles))

You can use this Python code to solve the given problem by taking input for the number of passengers and vehicles, as well as their coordinates, and then calculating the minimum distance traveled by all vehicles to reach their passengers based on the given rules.

Certainly! Here's an equivalent solution in C++ to tackle the Solo Rider's problem:

#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>

using namespace std;

// Structure to store coordinates
struct Point {
    string name;
    int x, y;
};

 Function to calculate Manhattan distance
int manhattanDistance(int x1, int y1, int2, int y2) {
    return abs(x1 - x2 + abs(y1 - y2);
}

// Function to find minimum distance traveled by all vehicles to reach their passengers
int minDistanceTraveled(int N, int M, vector<Point passengers, vector<Point>& vehicles) {
    int distance = 0;
    vector<string> assign(N,  // Stores the assignment of passengers to vehicles

    // passengers and vehicles based on lexicographical order of names and vehicle numbers
    sort(passengers.begin(), passengers.end(), [](const Point& a, const Point& b) {
        return a.name < b.name;
    });
    sort(vehicles.begin(), vehicles.end(), [](const Point& a, const Point& b) {
        return a.name <.name;
    });

    for (auto& passenger : passengers) {
        int minDist = INT_MAX;
        for (auto& vehicle : vehicles) {
            if (find(assign.begin(), assign.end(), vehicle.name) == assign.end()) {
                int dist = manhattanDistance(passenger.x, passenger.y, vehicle.x, vehicle.y);
                if (dist < minDist || (dist == minDist &&.name < assign[&passenger - &passengers[0]])) {
                    minDist = dist;
                    assign[&passenger - &passengers[0]] = vehicle.name;
                }
            }
        }
    }

    for (int i = 0; i < N; i++) {
        forauto& vehicle : vehicles) {
            if (vehicle.name == assign[i]) {
                distance += manhattanDistance(passengers[i].x, passengers[i].y, vehicle.x, vehicle.y);
                break;
            }
        }
    }

    return distance;
}

int main() {
    int N, M;
    cin >> N >> M;
    vector<Point> passengers(N), vehicles(M);

    for (int i = 0; < N; i++) {
        cin >> passengers[i].name >> passengers].x >> passengers[i].y;
    }
    for (int i = 0; i < M; i++) {
        cin >> vehicles[iname >> vehicles[i].x >> vehicles[i].y;
 }

    // Output
    cout << minDistanceTraveled(N, M, passengers, vehicles) << endl;

    return0;
}

You can use this C++ code address the Solo Rider's problem by taking input for the number of passengers and vehicles, as well as their coordinates, and then calculating the minimum distance traveled by all vehicles to reach their passengers based on the given rules.