检查一个数字在给定的基数内是否有连续的 0
原文:https://www . geesforgeks . org/check-一个数字在给定的基数上是否有连续的 0/
给定一个十进制数 N，任务是在将一个数转换为基于 K 的符号后，检查该数是否有连续的零。
示例:
输入:N = 4，K = 2 输出:否 2 中的 4 是 100，因为有连续的 2，所以答案是否
输入:N = 15，K = 8 输出:是 8 基数中的 15 为 17，由于没有连续的 0，所以答案为是。
逼近:先把数字 N 转换成基数 K，然后简单检查这个数字是否有连续的零。
下面是上述方法的实现:
C++
// C++ implementation of the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to convert N into base K
int toK(int N, int K)
{
// Weight of each digit
int w = 1;
int s = 0;
while (N != 0)
{
int r = N % K;
N = N/K;
s = r * w + s;
w *= 10;
}
return s;
}
// Function to check for consecutive 0
bool check(int N)
{
// Flag to check if there are consecutive
// zero or not
bool fl = false;
while (N != 0)
{
int r = N % 10;
N = N/10;
// If there are two consecutive zero
// then returning False
if (fl == true and r == 0)
return false;
if (r > 0)
{
fl = false;
continue;
}
fl = true;
}
return true;
}
// We first convert to given base, then
// check if the converted number has two
// consecutive 0s or not
void hasConsecutiveZeroes(int N, int K)
{
int z = toK(N, K);
if (check(z))
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
}
// Driver code
int main()
{
int N = 15;
int K = 8;
hasConsecutiveZeroes(N, K);
}
// This code is contributed by
// Surendra_Gangwar
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
import java.util.*;
class GFG
{
// Function to convert N into base K
static int toK(int N, int K)
{
// Weight of each digit
int w = 1;
int s = 0;
while (N != 0)
{
int r = N % K;
N = N / K;
s = r * w + s;
w *= 10;
}
return s;
}
// Function to check for consecutive 0
static boolean check(int N)
{
// Flag to check if there are consecutive
// zero or not
boolean fl = false;
while (N != 0)
{
int r = N % 10;
N = N / 10;
// If there are two consecutive zero
// then returning False
if (fl == true && r == 0)
return false;
if (r > 0)
{
fl = false;
continue;
}
fl = true;
}
return true;
}
// We first convert to given base, then
// check if the converted number has two
// consecutive 0s or not
static void hasConsecutiveZeroes(int N, int K)
{
int z = toK(N, K);
if (check(z))
System.out.println("Yes");
else
System.out.println("No");
}
// Driver code
public static void main(String[] args)
{
int N = 15;
int K = 8;
hasConsecutiveZeroes(N, K);
}
}
// This code is contributed by Princi Singh
Python 3
# Python implementation of the above approach
# We first convert to given base, then
# check if the converted number has two
# consecutive 0s or not
def hasConsecutiveZeroes(N, K):
z = toK(N, K)
if (check(z)):
print("Yes")
else:
print("No")
# Function to convert N into base K
def toK(N, K):
# Weight of each digit
w = 1
s = 0
while (N != 0):
r = N % K
N = N//K
s = r * w + s
w* = 10
return s
# Function to check for consecutive 0
def check(N):
# Flag to check if there are consecutive
# zero or not
fl = False
while (N != 0):
r = N % 10
N = N//10
# If there are two consecutive zero
# then returning False
if (fl == True and r == 0):
return False
if (r > 0):
fl = False
continue
fl = True
return True
# Driver code
N, K = 15, 8
hasConsecutiveZeroes(N, K)
C
// C# implementation of the above approach
using System;
class GFG
{
// Function to convert N into base K
static int toK(int N, int K)
{
// Weight of each digit
int w = 1;
int s = 0;
while (N != 0)
{
int r = N % K;
N = N / K;
s = r * w + s;
w *= 10;
}
return s;
}
// Function to check for consecutive 0
static Boolean check(int N)
{
// Flag to check if there are consecutive
// zero or not
Boolean fl = false;
while (N != 0)
{
int r = N % 10;
N = N / 10;
// If there are two consecutive zero
// then returning False
if (fl == true && r == 0)
return false;
if (r > 0)
{
fl = false;
continue;
}
fl = true;
}
return true;
}
// We first convert to given base, then
// check if the converted number has two
// consecutive 0s or not
static void hasConsecutiveZeroes(int N, int K)
{
int z = toK(N, K);
if (check(z))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
// Driver code
public static void Main(String[] args)
{
int N = 15;
int K = 8;
hasConsecutiveZeroes(N, K);
}
}
// This code is contributed by 29AjayKumar
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the above approach
// We first convert to given base,
// then check if the converted number
// has two consecutive 0s or not
function hasConsecutiveZeroes($N, $K)
{
$z = toK($N, $K);
if (check($z))
print("Yes");
else
print("No");
}
// Function to convert N into base K
function toK($N, $K)
{
// Weight of each digit
$w = 1;
$s = 0;
while ($N != 0)
{
$r = $N % $K;
$N = (int)($N / $K);
$s = $r * $w + $s;
$w *= 10;
}
return $s;
}
// Function to check for consecutive 0
function check($N)
{
// Flag to check if there are
// consecutive zero or not
$fl = false;
while ($N != 0)
{
$r = $N % 10;
$N = (int)($N / 10);
// If there are two consecutive
// zero then returning false
if ($fl == true and $r == 0)
return false;
if ($r > 0)
{
$fl = false;
continue;
}
$fl = true;
}
return true;
}
// Driver code
$N = 15;
$K = 8;
hasConsecutiveZeroes($N, $K);
// This code is contributed by mits
?>
java 描述语言
<script>
// Javascript implementation of the above approach
// Function to convert N into base K
function toK(N, K)
{
// Weight of each digit
let w = 1;
let s = 0;
while (N != 0)
{
let r = N % K;
N = parseInt(N / K);
s = r * w + s;
w *= 10;
}
return s;
}
// Function to check for consecutive 0
function check(N)
{
// Flag to check if there are consecutive
// zero or not
let fl = false;
while (N != 0)
{
let r = N % 10;
N = parseInt(N/10);
// If there are two consecutive zero
// then returning False
if (fl == true && r == 0)
return false;
if (r > 0)
{
fl = false;
continue;
}
fl = true;
}
return true;
}
// We first convert to given base, then
// check if the converted number has two
// consecutive 0s or not
function hasConsecutiveZeroes(N, K)
{
let z = toK(N, K);
if (check(z))
document.write("Yes");
else
document.write("No");
}
// Driver code
let N = 15;
let K = 8;
hasConsecutiveZeroes(N, K);
// This code is contributed by souravmahato348
</script>
Output:
Yes
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