通过重复乘以 10 或 20 来检查 N 是否可以从 1 中得到
原文:https://www . geesforgeks . org/check-if-n-can-from-1-by-reply-乘 10 或-20/
给定一个整数 N ，任务是通过重复乘以 10 或 20 来确定是否有可能从 1 获得值 N 。如果可能，打印“是”，否则打印“否”。 例:
输入: N = 200 输出: YES 解释: 1 * 10->10 * 20->200 输入: N = 90 输出: NO
进场: 按照以下步骤解决问题:
- 计算尾随零的数量。
- 去掉尾随零后，如果剩余的 N 不能表示为 2 的幂，则打印否。
- 否则，如果记录 2 N < =尾随零的计数，则打印“是”。
以下是上述方法的实施:
C++
// C++ program to check if N
// can be obtained from 1 by
// repetitive multiplication
// by 10 or 20
#include<bits/stdc++.h>
using namespace std;
// Function to check if N can
// be obtained or not
void Is_possible(long long int N)
{
int C = 0;
int D = 0;
// Count and remove trailing
// zeroes
while (N % 10 == 0)
{
N = N / 10;
C += 1;
}
// Check if remaining
// N is a power of 2
if(pow(2, (int)log2(N)) == N)
{
D = (int)log2(N);
// To check the condition
// to print YES or NO
if (C >= D)
cout << "YES";
else
cout << "NO";
}
else
cout << "NO";
}
// Driver code
int main()
{
long long int N = 2000000000000;
Is_possible(N);
}
// This code is contributed by Stream_Cipher
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to check if N
// can be obtained from 1 by
// repetitive multiplication
// by 10 or 20
import java.util.*;
class GFG{
static void Is_possible(long N)
{
long C = 0;
long D = 0;
// Count and remove trailing
// zeroes
while (N % 10 == 0)
{
N = N / 10;
C += 1;
}
// Check if remaining
// N is a power of 2
if(Math.pow(2, (long)(Math.log(N) /
(Math.log(2)))) == N)
{
D = (long)(Math.log(N) / (Math.log(2)));
// To check the condition
// to prlong YES or NO
if (C >= D)
System.out.print("YES");
else
System.out.print("NO");
}
else
System.out.print("NO");
}
// Driver code
public static void main(String args[])
{
long N = 2000000000000L;
Is_possible(N);
}
}
// This code is contributed by Stream_Cipher
计算机编程语言
# Python Program to check if N
# can be obtained from 1 by
# repetitive multiplication
# by 10 or 20
import math
# Function to check if N can
# be obtained or not
def Is_possible(N):
C = 0
D = 0
# Count and remove trailing
# zeroes
while ( N % 10 == 0):
N = N / 10
C += 1
# Check if remaining
# N is a power of 2
if ( math.log(N, 2)
- int(math.log(N, 2)) == 0):
D = int(math.log(N, 2))
# To check the condition
# to print YES or NO
if (C >= D):
print("YES")
else:
print("NO")
else:
print("NO")
# Driver Program
N = 2000000000000
Is_possible(N)
C
// C# program to check if N
// can be obtained from 1 by
// repetitive multiplication
// by 10 or 20
using System;
class GFG{
static void Is_possible(long N)
{
long C = 0;
long D = 0;
// Count and remove trailing
// zeroes
while (N % 10 == 0)
{
N = N / 10;
C += 1;
}
// Check if remaining
// N is a power of 2
if(Math.Pow(2, (long)(Math.Log(N) /
(Math.Log(2)))) == N)
{
D = (long)(Math.Log(N) / (Math.Log(2)));
// To check the condition
// to prlong YES or NO
if (C >= D)
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
else
Console.WriteLine("NO");
}
// Driver Code
public static void Main()
{
long N = 2000000000000L;
Is_possible(N);
}
}
// This code is contributed by Stream_Cipher
java 描述语言
<script>
// Java script program to check if N
// can be obtained from 1 by
// repetitive multiplication
// by 10 or 20
function Is_possible( N)
{
let C = 0;
let D = 0;
// Count and remove trailing
// zeroes
while (N % 10 == 0)
{
N = N / 10;
C += 1;
}
// Check if remaining
// N is a power of 2
if(Math.pow(2, (Math.log(N) /
(Math.log(2)))) == N)
{
D = (Math.log(N) / (Math.log(2)));
// To check the condition
// to prlong YES or NO
if (C >= D)
document.write("YES");
else
document.write("NO");
}
else
document.write("NO");
}
// Driver code
let N = 2000000000000;
Is_possible(N);
//this code is contributed by sravan kumar
</script>
Output:
YES
时间复杂度:O(log10(N)) 辅助空间: O(1)
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