检查 N 和 M 是否可以通过增加 A 的 N 和减少 B 的 M 而相等
原文:https://www . geeksforgeeks . org/check-if-n-and-m-can-make-equal-by-n-a-递增-n-a-递减-m-b/
给定四个数字 M 、 N 、 A 和 B ，任务是通过执行以下任一操作检查 M 和 N 是否可以相等:
- M 可增加 A 和 N 可减少 B
- 让他们两个保持原样。
示例:
输入: M = 2，N = 8，A = 3，B = 3 输出:是 说明: 第一次操作后: M 可以增加 A，所以 M = 2 + 3 = 5 N 可以减少 B，所以 N = 8–3 = 5 最后 M = N = 5。
输入: M = 6，N = 4，A = 2，B = 1 T3】输出:否
逼近:仔细观察可以发现，由于我们在增加 M，减少 N，所以只有 M 小于 N 时，才能使它们相等，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。
- 让他们两个保持原样。
可以做的另一个观察是当 M 增大而 N 减小时， M 和 N 之间的绝对距离减小了 A + B 的因子。例如:
设 M = 2，N = 14，A = 3，B = 3。
- 在步骤 1 中，M = 5，N = 11。M 和 N 之间的绝对距离减少了 6。也就是说，最初的绝对距离是 12(14–2)。执行给定步骤后，绝对距离变为 6(11–5)。
- 第二步，M = 8，N = 8。M 和 N 之间的绝对距离再次减少了 6，从而使 M 和 N 相等。
从上面的例子中，我们可以得出这样的结论:只有检查 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这个问题。
- 如果是倍数，那么 M 和 N 可以相等。
- 否则，他们就不能平等。
以下是上述方法的实现:
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 <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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_R
Python 3
# 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_R
C
// 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
Output:
Yes
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