检查数组表示的大数是否可以被 Y 整除
原文:https://www . geeksforgeeks . org/check-一个以数组形式表示的大数是否可被 y 整除/
给定一个大整数 X 表示为一个数组 arr[] ，其中每个 arr[i] 在 X 中存储一个数字。任务是检查数组表示的数字是否能被给定的整数 Y 整除。 举例:
输入: arr[] = {1，2，1，5，6}，Y = 4 输出:是 12156 / 4 = 3039 输入: arr[] = {1，1，1，1，1，1，1，1，1，1，Y = 14 输出:否
逼近:从左边开始遍历给定数字的数字，取小于等于 Y 的最大数字，用 Y 除。如果余数不是 0 那么它将被携带到由剩余数字形成的下一个可能的数字，就像在长除法中一样。处理完完整的数字后，如果余数仍然不是 0，则表示的数字不能被 Y 整除，否则就是 0。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function that returns true if the number represented
// by the given array is divisible by y
bool isDivisible(int* arr, int n, int y)
{
int d = 0, i = 0;
// While there are digits left
while (i < n) {
// Select the next part of the number
// i.e. the maximum number which is <= y
while (d < y && i < n)
d = d * 10 + arr[i++];
// Get the current remainder
d = d % y;
}
// If the final remainder is 0
if (d == 0)
return true;
return false;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 1, 5, 6 };
int x = sizeof(arr) / sizeof(int);
int y = 4;
cout << (isDivisible(arr, x, y) ? "Yes" : "No");
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function that returns true if the number represented
// by the given array is divisible by y
static boolean isDivisible(int [] arr, int n, int y)
{
int d = 0, i = 0;
// While there are digits left
while (i < n)
{
// Select the next part of the number
// i.e. the maximum number which is <= y
while (d < y && i < n)
d = d * 10 + arr[i++];
// Get the current remainder
d = d % y;
}
// If the final remainder is 0
if (d == 0)
return true;
return false;
}
// Driver code
public static void main (String[] args)
{
int [] arr = { 1, 2, 1, 5, 6 };
int x = arr.length;
int y = 4;
System.out.println(isDivisible(arr, x, y) ? "Yes" : "No");
}
}
// This code is contributed by ihritik
Python 3
# Python3 implementation of the approach
# Function that returns true if the number represented
# by the given array is divisible by y
def isDivisible(arr, n, y):
d, i = 0, 0
# While there are digits left
while i < n:
# Select the next part of the number
# i.e. the maximum number which is <= y
while d < y and i < n:
d = d * 10 + arr[i]
i += 1
# Get the current remainder
d = d % y
# If the final remainder is 0
if d == 0:
return True
return False
# Driver code
if __name__ == "__main__":
arr = [ 1, 2, 1, 5, 6 ]
x = len(arr)
y = 4
if (isDivisible(arr, x, y)):
print("Yes")
else:
print("No")
# This code is contributed by
# sanjeev2552
C
// C# implementation of the approach
using System;
class GFG
{
// Function that returns true if the number represented
// by the given array is divisible by y
static bool isDivisible(int [] arr, int n, int y)
{
int d = 0, i = 0;
// While there are digits left
while (i < n)
{
// Select the next part of the number
// i.e. the maximum number which is <= y
while (d < y && i < n)
d = d * 10 + arr[i++];
// Get the current remainder
d = d % y;
}
// If the final remainder is 0
if (d == 0)
return true;
return false;
}
// Driver code
public static void Main ()
{
int [] arr = { 1, 2, 1, 5, 6 };
int x = arr.Length;
int y = 4;
Console.WriteLine(isDivisible(arr, x, y) ? "Yes" : "No");
}
}
// This code is contributed by ihritik
java 描述语言
<script>
// JavaScript implementation of the approach
// Function that returns true if the number represented
// by the given array is divisible by y
function isDivisible(vararr , n , y)
{
var d = 0, i = 0;
// While there are digits left
while (i < n)
{
// Select the next part of the number
// i.e. the maximum number which is <= y
while (d < y && i < n)
d = d * 10 + arr[i++];
// Get the current remainder
d = d % y;
}
// If the final remainder is 0
if (d == 0)
return true;
return false;
}
// Driver code
var arr = [ 1, 2, 1, 5, 6 ];
var x = arr.length;
var y = 4;
document.write(isDivisible(arr, x, y) ? "Yes" : "No");
// This code is contributed by 29AjayKumar
</script>
Output:
Yes
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