给定两个整数X和Y ，任务是计算从点(1，1)开始到达点(1，1)所需的最小移动数(如果允许的话) ，例如(a， b)是(a – b，b)或(a，b – a)。
例子：
Input: X = 3, Y = 1
Output: 2
Explanation: Required sequence of moves are (3, 1) → (2, 1) → (1, 1)
Input: X = 2, Y = 2
Output: -1
天真的方法：解决问题的最简单方法是不断从较大的数中减去较小的数，直到它们相等为止。如果在任何时刻X或Y小于1 ，则打印-1 。否则，将执行的减法次数打印为所需的最小操作数。
时间复杂度： O(max(X，Y))
辅助空间： O(1)
高效方法：为了优化上述方法，其思想是通过取模运算来替换重复的减法，这与使用欧几里得算法计算GCD相似。请按照以下步骤解决问题：
- 初始化一个变量，例如用0表示cnt ，以存储最少的步骤数。
- 当X和Y都不为零时进行迭代，并执行以下操作：
- 如果X> Y的值，则将X / Y添加到cnt 。更新至X％Y
- 如果Y> X的值，则将Y / X添加到cnt 。更新y以Y％X
- 检查其中之一是否大于1 ，然后打印-1 。
- 否则，将cnt的值减小1，因为需要采取额外的步骤以使其中之一等于0 。
- 打印cnt的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to count the number of steps
// required to convert (x, y) to (1, 1)
int minimumSteps(int x, int y)
{
// Store the required result
int cnt = 0;
// Iterate while both x
// and y are not equal to 0
while (x != 0 && y != 0) {
// If x is greater than y
if (x > y) {
// Update count and value of x
cnt += x / y;
x %= y;
}
// Otherwise
else {
// Update count and value of y
cnt += y / x;
y %= x;
}
}
cnt--;
// If both x and y > 1
if (x > 1 || y > 1)
cnt = -1;
// Print the result
cout << cnt;
}
// Driver Code
int main()
{
// Given X and Y
int x = 3, y = 1;
minimumSteps(x, y);
return 0;
} Java
// Java program for above approach
import java.util.*;
class GFG{
// Function to count the number of steps
// required to convert (x, y) to (1, 1)
static void minimumSteps(int x, int y)
{
// Store the required result
int cnt = 0;
// Iterate while both x
// and y are not equal to 0
while (x != 0 && y != 0)
{
// If x is greater than y
if (x > y)
{
// Update count and value of x
cnt += x / y;
x %= y;
}
// Otherwise
else
{
// Update count and value of y
cnt += y / x;
y %= x;
}
}
cnt--;
// If both x and y > 1
if (x > 1 || y > 1)
cnt = -1;
// Print the result
System.out.println(cnt);
}
// Driver Code
public static void main (String[] args)
{
// Given X and Y
int x = 3, y = 1;
minimumSteps(x, y);
}
}
// This code is contributed by sanjoy_62Python3
# Python3 program for the above approach
# Function to count the number of steps
# required to convert (x, y) to (1, 1)
def minimumSteps(x, y):
# Store the required result
cnt = 0
# Iterate while both x
# and y are not equal to 0
while (x != 0 and y != 0):
# If x is greater than y
if (x > y):
# Update count and value of x
cnt += x / y
x %= y
# Otherwise
else:
# Update count and value of y
cnt += y / x
y %= x
cnt -= 1
# If both x and y > 1
if (x > 1 or y > 1):
cnt = -1
# Print the result
print(int(cnt))
# Driver Code
if __name__ == '__main__':
# Given X and Y
x = 3
y = 1
minimumSteps(x, y)
# This code is contributed by ipg2016107C#
// C# program for the above approach
using System;
class GFG{
// Function to count the number of steps
// required to convert (x, y) to (1, 1)
public static void minimumSteps(int x, int y)
{
// Store the required result
int cnt = 0;
// Iterate while both x
// and y are not equal to 0
while (x != 0 && y != 0)
{
// If x is greater than y
if (x > y)
{
// Update count and value of x
cnt += x / y;
x %= y;
}
// Otherwise
else
{
// Update count and value of y
cnt += y / x;
y %= x;
}
}
cnt--;
// If both x and y > 1
if (x > 1 || y > 1)
cnt = -1;
// Print the result
Console.WriteLine(cnt);
}
// Driver Code
public static void Main()
{
// Given X and Y
int x = 3, y = 1;
minimumSteps(x, y);
}
}
// This code is contributed by mohit kumar 29输出：
2时间复杂度： O(log(max(X(Y，Y))))
辅助空间： O(1)