给定两个整数n和m ,在一次运算中, n可以乘以2或3 。任务是用最少的给定操作数将n转换为m 。如果无法通过给定的操作将n转换为m ,则打印-1 。
例子:
Input: n = 120, m = 51840
Output: 7
120 * 2 * 2 * 2 * 2 * 3 * 3 * 3 = 51840
Input: n = 42, m = 42
Output: 0
No operation required.
Input: n = 48, m = 72
Output: -1
方法:如果m不能被n整除,则打印-1,因为无法通过给定的操作将n转换为m 。另外,我们可以检查商除以商时是否只有2和3作为主要因子。如果是,那么结果将是2和3的幂的和,否则打印-1
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the minimum
// operations required
int minOperations(int n, int m)
{
if (m % n != 0)
return -1;
int minOperations = 0;
int q = m / n;
// Counting all 2s
while (q % 2 == 0) {
q = q / 2;
minOperations++;
}
// Counting all 3s
while (q % 3 == 0) {
q = q / 3;
minOperations++;
}
// If q contained only 2 and 3
// as the only prime factors
// then it must be 1 now
if (q == 1)
return minOperations;
return -1;
}
// Driver code
int main()
{
int n = 120, m = 51840;
cout << minOperations(n, m);
return 0;
}
Java
// Java implementation of the approach
class GfG {
// Function to return the minimum
// operations required
static int minOperations(int n, int m)
{
if (m % n != 0)
return -1;
int minOperations = 0;
int q = m / n;
// Counting all 2s
while (q % 2 == 0) {
q = q / 2;
minOperations++;
}
// Counting all 3s
while (q % 3 == 0) {
q = q / 3;
minOperations++;
}
// If q contained only 2 and 3
// as the only prime factors
// then it must be 1 now
if (q == 1)
return minOperations;
return -1;
}
// Driver code
public static void main(String[] args)
{
int n = 120, m = 51840;
System.out.println(minOperations(n, m));
}
}
Python3
# Python 3 implementation of the approach
# Function to return the minimum
# operations required
def minOperations(n, m):
if (m % n != 0):
return -1
minOperations = 0
q = int(m / n)
# Counting all 2s
while (q % 2 == 0):
q = int(q / 2)
minOperations += 1
# Counting all 3s
while (q % 3 == 0):
q = int(q / 3)
minOperations += 1
# If q contained only 2 and 3
# as the only prime factors
# then it must be 1 now
if (q == 1):
return minOperations
return -1
# Driver code
if __name__ == '__main__':
n = 120
m = 51840
print(minOperations(n, m))
# This code is contributed by
# Surendra_Gangwar
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the minimum
// operations required
static int minOperations(int n, int m)
{
if (m % n != 0)
return -1;
int minOperations = 0;
int q = m / n;
// Counting all 2s
while (q % 2 == 0)
{
q = q / 2;
minOperations++;
}
// Counting all 3s
while (q % 3 == 0)
{
q = q / 3;
minOperations++;
}
// If q contained only 2 and 3
// as the only prime factors
// then it must be 1 now
if (q == 1)
return minOperations;
return -1;
}
// Driver code
public static void Main()
{
int n = 120, m = 51840;
Console.WriteLine(minOperations(n, m));
}
}
// This code is contributed
// by Akanksha Rai
PHP
Javascript
输出:
7
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