给定四个数字M ， N ， A和B ，任务是通过执行以下任一操作来检查M和N是否可以彼此相等：
- M可以增加A ， N可以减少B
- 保持原样。
例子：
Input: M = 2, N = 8, A = 3, B = 3
Output: Yes
Explanation:
After first Operation:
M can be increased by A. Therefore, M = 2 + 3 = 5
N can be decreased by B. Therefore, N = 8 – 3 = 5
Finally, M = N = 5.
Input: M = 6, N = 4, A = 2, B = 1
Output: No
方法：仔细观察，可以发现由于我们增加M而减少N，因此只有当M小于N时，它们才能相等。因此，当M小于N时，每一步都有两种情况：
- M可以增加A，N可以减少B。
- 保持原样。
可以得出的另一个观察结果是，当M增加而N减小时， M与N之间的绝对距离减小A + B倍。例如：
Let M = 2, N = 14, A = 3 and B = 3.
- In step 1, M = 5 and N = 11. The absolute distance between M and N got reduced by 6. That is, initially, the absolute distance was 12(14 – 2). After performing the given step, the absolute distance became 6(11 – 5).
- In step 2, M = 8 and N = 8. The absolute distance between M and N again got reduced by 6 thereby making M and N equal.
从以上示例可以得出结论，只有通过检查M和N之间的绝对距离是否为(A + B)的倍数，才能在恒定时间内解决此问题。
- 如果是倍数，则M和N可以相等。
- 否则，它们不能相等。
下面是上述方法的实现：
C++
// C++ program to check if two numbers
// can be made equal by increasing
// the first by a and decreasing
// the second by b
#include
using namespace std;
// Function to whether the numbers
// can be made equal or not
bool checkEqualNo(int m, int n, int a, int b)
{
if (m <= n) {
// Check whether the numbers can reach
// an equal point or not
if ((n - m) % (a + b) == 0) {
return true;
}
else {
return false;
}
}
else {
// M and N cannot be made equal by
// increasing M and decreasing N when
// M is already greater than N
return false;
}
}
// Driver code
int main()
{
int M = 2, N = 8;
int A = 3, B = 3;
if (checkEqualNo(M, N, A, B))
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
} Java
// Java program to check if two numbers
// can be made equal by increasing
// the first by a and decreasing
// the second by b
class GFG
{
// Function to whether the numbers
// can be made equal or not
static boolean checkEqualNo(int m, int n, int a, int b)
{
if (m <= n) {
// Check whether the numbers can reach
// an equal point or not
if ((n - m) % (a + b) == 0) {
return true;
}
else {
return false;
}
}
else {
// M and N cannot be made equal by
// increasing M and decreasing N when
// M is already greater than N
return false;
}
}
// Driver code
public static void main (String[] args)
{
int M = 2, N = 8;
int A = 3, B = 3;
if (checkEqualNo(M, N, A, B) == true)
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Yash_RPython3
# Python3 program to check if two numbers
# can be made equal by increasing
# the first by a and decreasing
# the second by b
# Function to whether the numbers
# can be made equal or not
def checkEqualNo(m, n, a, b) :
if (m <= n) :
# Check whether the numbers can reach
# an equal point or not
if ((n - m) % (a + b) == 0) :
return True;
else :
return False;
else :
# M and N cannot be made equal by
# increasing M and decreasing N when
# M is already greater than N
return False;
# Driver code
if __name__ == "__main__" :
M = 2; N = 8;
A = 3; B = 3;
if (checkEqualNo(M, N, A, B)) :
print("Yes");
else :
print("No");
# This code is contributed by Yash_RC#
// C# program to check if two numbers
// can be made equal by increasing
// the first by a and decreasing
// the second by b
using System;
class GFG
{
// Function to whether the numbers
// can be made equal or not
static bool checkEqualNo(int m, int n, int a, int b)
{
if (m <= n) {
// Check whether the numbers can reach
// an equal point or not
if ((n - m) % (a + b) == 0) {
return true;
}
else {
return false;
}
}
else {
// M and N cannot be made equal by
// increasing M and decreasing N when
// M is already greater than N
return false;
}
}
// Driver code
public static void Main (String[] args)
{
int M = 2;
int N = 8;
int A = 3;
int B = 3;
if (checkEqualNo(M, N, A, B) == true)
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by Yash_R输出：
Yes