具有整数交点的线对数

给定两个整数数组P[]和Q[] ，其中p _i和q _j对于每个0 <= i < size(P)和0 <= j < size(Q)表示线方程x – y = -p _i和x + y = q _j分别。任务是从具有整数交点的P[]和Q[]中找到对的数量。

例子：

Input: P[] = {1, 3, 2}, Q[] = {3, 0} 
Output: 3 
The pairs of lines (p, q) having integer intersection points are (1, 3), (2, 0) and (3, 3). Here p is the line parameter of P[] and q is the that of Q[].

Input: P[] = {1, 4, 3, 2}, Q[] = {3, 6, 10, 11} 
Output: 8 
 

方法：

• 通过求解两个方程并分析整数交点的条件，可以轻松解决该问题。
• 这两个方程是x – y = -p和x + y = q 。
• 求解x和y我们得到， x = (qp)/2和y = (p+q)/2 。
• 很明显，当且仅当p和q具有相同的奇偶性时，整数交点是可能的。
• 设p ₀和p ₁分别是偶数和奇数p _i 。
• 类似地， q ₀和q _{1 分别}为偶数和奇数q _i 。
• 因此所需的答案是p ₀ * q ₀ + p ₁ * q ₁ 。

下面是上述方法的实现：

C++

// C++ program to Number of pairs of lines
// having integer intersection points
 
#include 
using namespace std;
 
// Count number of pairs of lines
// having integer intersection point
int countPairs(int* P, int* Q, int N, int M)
{
    // Initialize arrays to store counts
    int A[2] = { 0 }, B[2] = { 0 };
 
    // Count number of odd and even Pi
    for (int i = 0; i < N; i++)
        A[P[i] % 2]++;
 
    // Count number of odd and even Qi
    for (int i = 0; i < M; i++)
        B[Q[i] % 2]++;
 
    // Return the count of pairs
    return (A[0] * B[0] + A[1] * B[1]);
}
 
// Driver code
int main()
{
    int P[] = { 1, 3, 2 }, Q[] = { 3, 0 };
    int N = sizeof(P) / sizeof(P[0]);
    int M = sizeof(Q) / sizeof(Q[0]);
 
    cout << countPairs(P, Q, N, M);
 
    return 0;
}

Java

// Java program to Number of pairs of lines
// having integer intersection points
class GFG
{
 
// Count number of pairs of lines
// having integer intersection point
static int countPairs(int []P, int []Q,
                      int N, int M)
{
    // Initialize arrays to store counts
    int []A = new int[2], B = new int[2];
 
    // Count number of odd and even Pi
    for (int i = 0; i < N; i++)
        A[P[i] % 2]++;
 
    // Count number of odd and even Qi
    for (int i = 0; i < M; i++)
        B[Q[i] % 2]++;
 
    // Return the count of pairs
    return (A[0] * B[0] + A[1] * B[1]);
}
 
// Driver code
public static void main(String[] args)
{
    int []P = { 1, 3, 2 };
    int []Q = { 3, 0 };
    int N = P.length;
    int M = Q.length;
 
    System.out.print(countPairs(P, Q, N, M));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to Number of pairs of lines
# having eger ersection pos
 
# Count number of pairs of lines
# having eger ersection po
def countPairs(P, Q, N, M):
     
    # Initialize arrays to store counts
    A = [0] * 2
    B = [0] * 2
 
    # Count number of odd and even Pi
    for i in range(N):
        A[P[i] % 2] += 1
 
    # Count number of odd and even Qi
    for i in range(M):
        B[Q[i] % 2] += 1
 
    # Return the count of pairs
    return (A[0] * B[0] + A[1] * B[1])
 
# Driver code
 
P = [1, 3, 2]
Q = [3, 0]
N = len(P)
M = len(Q)
 
print(countPairs(P, Q, N, M))
 
# This code is contributed by mohit kumar 29

C#

// C# program to Number of pairs of lines
// having integer intersection points
using System;
 
class GFG
{
     
    // Count number of pairs of lines
    // having integer intersection point
    static int countPairs(int []P, int []Q,
                        int N, int M)
    {
        // Initialize arrays to store counts
        int []A = new int[2];
        int []B = new int[2];
     
        // Count number of odd and even Pi
        for (int i = 0; i < N; i++)
            A[P[i] % 2]++;
     
        // Count number of odd and even Qi
        for (int i = 0; i < M; i++)
            B[Q[i] % 2]++;
     
        // Return the count of pairs
        return (A[0] * B[0] + A[1] * B[1]);
    }
     
    // Driver code
    public static void Main()
    {
        int []P = { 1, 3, 2 };
        int []Q = { 3, 0 };
        int N = P.Length;
        int M = Q.Length;
     
        Console.Write(countPairs(P, Q, N, M));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

 
// Javascript program to Number of
// pairs of lines having integer
// intersection points
 
// Count number of pairs of lines
// having integer intersection point
function countPairs(P, Q, N, M)
{
     
    // Initialize arrays to store counts
    var A = [0, 0], B = [0, 0];
 
    // Count number of odd and even Pi
    for(var i = 0; i < N; i++)
        A[P[i] % 2]++;
 
    // Count number of odd and even Qi
    for(var i = 0; i < M; i++)
        B[Q[i] % 2]++;
 
    // Return the count of pairs
    return(A[0] * B[0] + A[1] * B[1]);
}
 
// Driver code
var P = [ 1, 3, 2 ], Q = [ 3, 0 ];
var N = P.length;
var M = Q.length;
 
document.write(countPairs(P, Q, N, M));
 
// This code is contributed by rrrtnx
 

输出：

3

时间复杂度： O(P + Q)