x <y的解数，其中a <= x <= b和c <= y <= d并且x，y是整数

给定四个整数a，b，c，d(最大为10 ^ 6)。任务是找到x

例子：

Input: a = 2, b = 3, c = 3, d = 4
Output: 3

Input: a = 3, b = 5, c = 6, d = 7
Output: 6

方法：让我们显式地迭代x的所有可能值。对于x的一个这样的固定值，问题减少到存在多少y值，使得c <= y <= d且x = max(c，x + 1)且y <= d 。假定c <= d ，否则，当然没有y的有效值。因此，对于固定的x，存在d – max(c，x + 1)+ 1个y的有效值，因为在[R1，R2]范围内的整数数目由R2 – R1 + 1给出。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
  
// function to Find the number of solutions for x < y,
// where a <= x <= b and c <= y <= d and x, y integers.
int NumberOfSolutions(int a, int b, int c, int d)
{
    // to store answer
    int ans = 0;
  
    // iterate explicitly over all possible values of x
    for (int i = a; i <= b; i++)
        if (d >= max(c, i + 1))
            ans += d - max(c, i + 1) + 1;
  
    // return answer
    return ans;
}
  
// Driver code
int main()
{
    int a = 2, b = 3, c = 3, d = 4;
  
    // function call
    cout << NumberOfSolutions(a, b, c, d);
  
    return 0;
}

Java

// Java implementation of above approach
import java.io.*;
  
class GFG 
{
  
// function to Find the number of
// solutions for x < y, where
// a <= x <= b and c <= y <= d 
// and x, y integers.
static int NumberOfSolutions(int a, int b, 
                             int c, int d)
{
    // to store answer
    int ans = 0;
  
    // iterate explicitly over all 
    // possible values of x
    for (int i = a; i <= b; i++)
        if (d >= Math.max(c, i + 1))
            ans += d - Math.max(c, i + 1) + 1;
  
    // return answer
    return ans;
}
  
// Driver code
public static void main (String[] args) 
{
    int a = 2, b = 3, c = 3, d = 4;
  
    // function call
    System.out.println(NumberOfSolutions(a, b, c, d));
}
}
  
// This code is contributed 
// by inder_verma

Python 3

# Python3 implementation of 
# above approach
  
# function to Find the number of 
# solutions for x < y, where 
# a <= x <= b and c <= y <= d and
# x, y integers. 
def NumberOfSolutions(a, b, c, d) :
  
    # to store answer 
    ans = 0
  
    # iterate explicitly over all 
    # possible values of x 
    for i in range(a, b + 1) :
  
        if d >= max(c, i + 1) :
  
            ans += d - max(c, i + 1) + 1
  
    # return answer 
    return ans
  
# Driver code
if __name__ == "__main__" :
  
    a, b, c, d = 2, 3, 3, 4
  
    # function call 
    print(NumberOfSolutions(a, b, c, d))
  
# This code is contributed by ANKITRAI1

C#

// C# implementation of above approach
using System;
  
class GFG 
{
  
// function to Find the number of
// solutions for x < y, where
// a <= x <= b and c <= y <= d 
// and x, y integers.
static int NumberOfSolutions(int a, int b, 
                              int c, int d)
{
    // to store answer
    int ans = 0;
  
    // iterate explicitly over all 
    // possible values of x
    for (int i = a; i <= b; i++)
        if (d >= Math.Max(c, i + 1))
            ans += d - Math.Max(c, i + 1) + 1;
  
    // return answer
    return ans;
}
  
// Driver code
public static void Main() 
{
    int a = 2, b = 3, c = 3, d = 4;
  
    // function call
    Console.WriteLine(NumberOfSolutions(a, b, c, d));
}
}
  
// This code is contributed 
// by Akanksha Rai(Abby_akku)

PHP

= max($c, $i + 1))
            $ans += $d - max($c, $i + 1) + 1;
  
    // return answer
    return $ans;
}
  
// Driver code
$a = 2; $b = 3; $c = 3; $d = 4;
  
// function call
echo NumberOfSolutions($a, $b, $c, $d);
  
// This code is contributed
// by Akanksha Rai(Abby_akku)
?>

输出：

3