两列火车在同一方向行驶的距离

给定两个数组A[]和B[] ，每个数组由N 个整数组成，包含两列火车在同一方向行驶的速度，在每个时间单位，任务是找到两列火车一起行驶的总距离(边并排)在整个旅程中。

例子：

Input: A[] = {1, 2, 3, 2, 4}, B[] = {2, 1, 3, 1, 4} 
Output: 3 
Explanation : 
Since A[1] + A[0] = B[0] + B[1], both the trains have travelled same distance after 2 units of time. 
Now, since A[2] = B[2] = 3, both the trains have traveled this distance together. 
After the 3rd unit of time, the speed of the trains are different. 
Therefore, the total distance traveled by the two trains together is 3.
Input: A[] = {1, 1, 3, 2, 4}, B[] = {3, 1, 2, 1, 4} 
Output:

方法：
请按照以下步骤解决问题：

• 同时遍历两个数组。
• 对于每个^第i^个索引，检查sum(A[0] .. A[i – 1])是否等于sum(B[0] .. B[i – 1])以及A[i]和B [i]是否相等。
• 如果满足以上两个条件，则将A[i]添加到答案中。
• 最后，遍历完整个数组后，打印answer 。

下面是上述方法的实现：

C++

// C++ Program to find the distance
// traveled together by the two trains
#include 
using namespace std;
 
// Function to find the distance traveled together
int calc_distance(int A[], int B[], int n)
{
 
    // Stores distance travelled by A
    int distance_traveled_A = 0;
 
    // Stpres distance travelled by B
    int distance_traveled_B = 0;
 
    // Stores the total distance
    // travelled together
    int answer = 0;
 
    for (int i = 0; i < 5; i++) {
 
        // Sum of distance travelled
        distance_traveled_A += A[i];
        distance_traveled_B += B[i];
 
        // Condition for traveling
        // together
        if ((distance_traveled_A
             == distance_traveled_B)
            && (A[i] == B[i])) {
            answer += A[i];
        }
    }
    return answer;
}
 
// Driver Code
int main()
{
 
    int A[5] = { 1, 2, 3, 2, 4 };
    int B[5] = { 2, 1, 3, 1, 4 };
    int N = sizeof(A) / sizeof(A[0]);
 
    cout << calc_distance(A, B, N);
    return 0;
}

Java

// Java program to find the distance
// traveled together by the two trains
import java.util.*;
import java.lang.*;
 
class GFG{
     
// Function to find the distance traveled together
static int calc_distance(int A[], int B[], int n)
{
 
    // Stores distance travelled by A
    int distance_traveled_A = 0;
 
    // Stpres distance travelled by B
    int distance_traveled_B = 0;
 
    // Stores the total distance
    // travelled together
    int answer = 0;
 
    for(int i = 0; i < 5; i++)
    {
 
        // Sum of distance travelled
        distance_traveled_A += A[i];
        distance_traveled_B += B[i];
 
        // Condition for traveling
        // together
        if ((distance_traveled_A ==
             distance_traveled_B) &&
             (A[i] == B[i]))
        {
            answer += A[i];
        }
    }
    return answer;
}
 
// Driver code
public static void main (String[] args)
{
    int A[] = { 1, 2, 3, 2, 4 };
    int B[] = { 2, 1, 3, 1, 4 };
    int N = A.length;
     
    System.out.println(calc_distance(A, B, N));    
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to find the distance
# traveled together by the two trains
 
# Function to find the distance
# traveled together
def calc_distance(A, B, n):
 
    # Stores distance travelled by A
    distance_traveled_A = 0
 
    # Stpres distance travelled by B
    distance_traveled_B = 0
 
    # Stores the total distance
    # travelled together
    answer = 0
 
    for i in range(5):
 
        # Sum of distance travelled
        distance_traveled_A += A[i]
        distance_traveled_B += B[i]
 
        # Condition for traveling
        # together
        if ((distance_traveled_A ==
             distance_traveled_B) and
            (A[i] == B[i])):
            answer += A[i]
     
    return answer
 
# Driver Code
A = [ 1, 2, 3, 2, 4 ]
B = [ 2, 1, 3, 1, 4 ]
 
N = len(A)
 
print(calc_distance(A, B, N))
 
# This code is contributed by sanjoy_62

C#

// C# program to find the distance
// traveled together by the two trains
using System;
 
class GFG{
 
// Function to find the distance
// traveled together
static int calc_distance(int []A, int []B,
                         int n)
{
 
    // Stores distance travelled by A
    int distance_traveled_A = 0;
 
    // Stpres distance travelled by B
    int distance_traveled_B = 0;
 
    // Stores the total distance
    // travelled together
    int answer = 0;
 
    for(int i = 0; i < 5; i++)
    {
 
        // Sum of distance travelled
        distance_traveled_A += A[i];
        distance_traveled_B += B[i];
 
        // Condition for traveling
        // together
        if ((distance_traveled_A ==
             distance_traveled_B) &&
             (A[i] == B[i]))
        {
            answer += A[i];
        }
    }
    return answer;
}
 
// Driver Code
public static void Main(string []s)
{
    int []A = { 1, 2, 3, 2, 4 };
    int []B = { 2, 1, 3, 1, 4 };
    int N = A.Length;
 
    Console.Write(calc_distance(A, B, N));
}
}
 
// This code is contributed by rutvik_56

Javascript

// javascript program to find the distance
// traveled together by the two trains
// Function to find the distance traveled together
    function calc_distance(A , B , n)
    {
 
        // Stores distance travelled by A
        var distance_traveled_A = 0;
 
        // Stpres distance travelled by B
        var distance_traveled_B = 0;
 
        // Stores the total distance
        // travelled together
        var answer = 0;
 
        for (i = 0; i < 5; i++) {
 
            // Sum of distance travelled
            distance_traveled_A += A[i];
            distance_traveled_B += B[i];
 
            // Condition for traveling
            // together
            if ((distance_traveled_A == distance_traveled_B) && (A[i] == B[i])) {
                answer += A[i];
            }
        }
        return answer;
    }
 
    // Driver code
     
        var A = [ 1, 2, 3, 2, 4 ];
        var B = [ 2, 1, 3, 1, 4 ];
        var N = A.length;
 
        document.write(calc_distance(A, B, N));
 
// This code is contributed by aashish1995

输出：

3

时间复杂度： O(N)
辅助空间： O(1)