计算给定条件下的最大功率
给定 3 个整数 N，M ，和 K 。任务是找到最大 P 使得 N * K P ≤ M 。
示例:
输入: N = 1，K = 2，M = 5 输出: 2 输入: N = 5，K = 25，M = 100 输出: 0
逼近: 在该算法中，只需将 N 与 K 相乘，并用结果更新 N 的当前值，并将可变功率(初始为 0)增加 1 。 为了实现这一点，定义了一个递归函数，它有两种基本情况。
- 最初，N 大于要求，因此，返回 0 。
-
否则，返回电源–1。
-
如果当前 N 值等于 M，则返回功率。
- 递归条件如果当前 N < M : 更新 N 为 (N * k) ，功率为当前功率+ 1 。
以下是上述方法的实现:
C++
// Compute maximum power to which K can be raised so
// that given condition remains true
#include <bits/stdc++.h>
using namespace std;
#define ll long long
// Function to return the largest
// power
int calculate(ll int n, ll int k,
ll int m, ll int power)
{
// If n is greater than given M
if (n > m) {
if (power == 0)
return 0;
else
return power - 1;
}
// If n == m
else if (n == m)
return power;
else
// Checking for the next power
return calculate(n * k, k, m, power + 1);
}
// Driver Code
int main()
{
ll N = 1, K = 2, M = 5;
cout << calculate(N, K, M, 0);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for Compute maximum power
// to which K can be raised so that
// given condition remains true
class GFG
{
// Function to return the largest
// power
static int calculate(int n, int k,
int m, int power)
{
// If n is greater than given M
if (n > m)
{
if (power == 0)
return 0;
else
return power - 1;
}
// If n == m
else if (n == m)
return power;
else
// Checking for the next power
return calculate(n * k, k, m,
power + 1);
}
// Driver Code
public static void main (String[] args)
{
int N = 1, K = 2, M = 5;
System.out.println(calculate(N, K, M, 0));
}
}
// This code is contributed by AnkitRai01
计算机编程语言
# Compute maximum power to
# which K can be raised so
# that given condition
# remains true
# Function to return the largest
# power
def calculate(n, k, m, power):
# If n is greater than given M
if n > m:
if power == 0:
return 0
else:
return power-1
# If n == m
elif n == m:
return power
else:
# Checking for the next power
return calculate(n * k, k, m, power + 1)
# Driver's code
if __name__=="__main__":
N = 1
K = 2
M = 5
print(calculate(N, K, M, 0))
C
// C# program for Compute maximum power
// to which K can be raised so that
// given condition remains true
using System;
class GFG
{
// Function to return the largest
// power
static int calculate(int n, int k,
int m, int power)
{
// If n is greater than given M
if (n > m)
{
if (power == 0)
return 0;
else
return power - 1;
}
// If n == m
else if (n == m)
return power;
else
// Checking for the next power
return calculate(n * k, k, m,
power + 1);
}
// Driver Code
public static void Main (String[] args)
{
int N = 1, K = 2, M = 5;
Console.WriteLine(calculate(N, K, M, 0));
}
}
// This code is contributed by PrinciRaj1992
java 描述语言
<script>
// javascript program for Compute maximum power
// to which K can be raised so that
// given condition remains true
// Function to return the largest
// power
function calculate(n , k , m , power) {
// If n is greater than given M
if (n > m) {
if (power == 0)
return 0;
else
return power - 1;
}
// If n == m
else if (n == m)
return power;
else
// Checking for the next power
return calculate(n * k, k, m, power + 1);
}
// Driver Code
var N = 1, K = 2, M = 5;
document.write(calculate(N, K, M, 0));
// This code contributed by aashish1995
</script>
Output:
2
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