给定一个包含n个数字的数组。任务是将所有n个数字相乘后，在末尾打印连续零的数量。 例如：
null
Input : arr[] = {100, 10}Output : 3Explanation : 100 x 10 = 1000,3 zero's at the end.Input : arr[] = {100, 10, 5, 25, 35, 14}Output : 4Explanation :100 x 10 x 5 x 25 x 35 x 14 = 61250000,4 zero's at the end
天真的方法： 首先将所有数字相乘并保存到字符串中（因为如果乘法数大到2^64，那么它给出的是错误的，所以将每个倍数存储在字符串中），然后计算零的数量。 有效方法： 首先计算n个数字中的2的因子，然后计算所有n个数字中的5的因子，然后打印最小的一个。 例如——
n_number's | 2's factor | 5's factor 100 | 2 | 210 | 1 | 15 | 0 | 125 | 0 | 235 | 0 | 1 14 | 1 | 0Total | 4 | 7we can take a minimum so there number of zero's is 4
以下是上述方法的实施情况：
C++
// CPP program to find the number of consecutive zero // at the end after multiplying n numbers #include<iostream> using namespace std; // Function to count two's factor int two_factor( int n) { // Count number of 2s present in n int twocount = 0; while (n % 2 == 0) { twocount++; n = n / 2; } return twocount; } // Function to count five's factor int five_factor( int n) { int fivecount = 0; while (n % 5 == 0) { fivecount++; n = n / 5; } return fivecount; } // Function to count number of zeros int find_con_zero( int arr[], int n) { int twocount = 0; int fivecount = 0; for ( int i = 0; i < n; i++) { // Count the two's factor of n number twocount += two_factor(arr[i]); // Count the five's factor of n number fivecount += five_factor(arr[i]); } // Return the minimum if (twocount < fivecount) return twocount; else return fivecount; } // Driver Code int main() { int arr[] = { 100, 10, 5, 25, 35, 14 }; int n = 6; cout << find_con_zero(arr, n); } |
JAVA
// Java program to find the number // of consecutive zero at the end // after multiplying n numbers public class GfG{ // Function to count two's factor static int two_factor( int n) { // Count number of 2s // present in n int twocount = 0 ; while (n % 2 == 0 ) { twocount++; n = n / 2 ; } return twocount; } // Function to count five's // factor static int five_factor( int n) { int fivecount = 0 ; while (n % 5 == 0 ) { fivecount++; n = n / 5 ; } return fivecount; } // Function to count number of zeros static int find_con_zero( int arr[], int n) { int twocount = 0 ; int fivecount = 0 ; for ( int i = 0 ; i < n; i++) { // Count the two's factor // of n number twocount += two_factor(arr[i]); // Count the five's factor // of n number fivecount += five_factor(arr[i]); } // Return the minimum if (twocount < fivecount) return twocount; else return fivecount; } // driver function public static void main(String s[]) { int arr[] = { 100 , 10 , 5 , 25 , 35 , 14 }; int n = 6 ; System.out.println(find_con_zero(arr, n)); } } // This code is contributed by Gitanjali |
Python3
# Python3 code to find the number of consecutive zero # at the end after multiplying n numbers # Function to count two's factor def two_factor( n ): # Count number of 2s present in n twocount = 0 while n % 2 = = 0 : twocount + = 1 n = int ( n / 2 ) return twocount # Function to count five's factor def five_factor( n ): fivecount = 0 while n % 5 = = 0 : fivecount + = 1 n = int (n / 5 ) return fivecount # Function to count number of zeros def find_con_zero( arr, n ): twocount = 0 fivecount = 0 for i in range (n): # Count the two's factor of n number twocount + = two_factor(arr[i]) # Count the five's factor of n number fivecount + = five_factor(arr[i]) # Return the minimum if twocount < fivecount: return twocount else : return fivecount # Driver Code arr = [ 100 , 10 , 5 , 25 , 35 , 14 ] n = 6 print (find_con_zero(arr, n)) # This code is contributed by "Sharad_Bhardwaj". |
C#
// C# program to find the number // of consecutive zero at the end // after multiplying n numbers using System; public class GfG { // Function to count two's factor static int two_factor( int n) { // Count number of 2s // present in n int twocount = 0; while (n % 2 == 0) { twocount++; n = n / 2; } return twocount; } // Function to count five's // factor static int five_factor( int n) { int fivecount = 0; while (n % 5 == 0) { fivecount++; n = n / 5; } return fivecount; } // Function to count number of zeros static int find_con_zero( int []arr, int n) { int twocount = 0; int fivecount = 0; for ( int i = 0; i < n; i++) { // Count the two's factor // of n number twocount += two_factor(arr[i]); // Count the five's factor // of n number fivecount += five_factor(arr[i]); } // Return the minimum if (twocount < fivecount) return twocount; else return fivecount; } // driver function public static void Main() { int []arr = { 100, 10, 5, 25, 35, 14 }; int n = 6; Console.WriteLine(find_con_zero(arr, n)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to find the number // of consecutive zero at the end // after multiplying n numbers // Function to count two's factor function two_factor( $n ) { // Count number of // 2s present in n $twocount = 0; while ( $n % 2 == 0) { $twocount ++; $n = (int)( $n / 2); } return $twocount ; } // Function to count // five's factor function five_factor( $n ) { $fivecount = 0; while ( $n % 5 == 0) { $fivecount ++; $n = (int)( $n / 5); } return $fivecount ; } // Function to count // number of zeros function find_con_zero( $arr , $n ) { $twocount = 0; $fivecount = 0; for ( $i = 0; $i < $n ; $i ++) { // Count the two's // factor of n number $twocount += two_factor( $arr [ $i ]); // Count the five's // factor of n number $fivecount += five_factor( $arr [ $i ]); } // Return the minimum if ( $twocount < $fivecount ) return $twocount ; else return $fivecount ; } // Driver Code $arr = array (100, 10, 5, 25, 35, 14); $n = 6; echo find_con_zero( $arr , $n ); // This code is contributed by mits ?> |
Javascript
<script> // JavaScript program to find the number // of consecutive zero at the end // after multiplying n numbers // Function to count two's factor function two_factor(n) { // Count number of 2s // present in n let twocount = 0; while (n % 2 == 0) { twocount++; n = n / 2; } return twocount; } // Function to count five's // factor function five_factor(n) { let fivecount = 0; while (n % 5 == 0) { fivecount++; n = n / 5; } return fivecount; } // Function to count number of zeros function find_con_zero(arr, n) { let twocount = 0; let fivecount = 0; for (let i = 0; i < n; i++) { // Count the two's factor // of n number twocount += two_factor(arr[i]); // Count the five's factor // of n number fivecount += five_factor(arr[i]); } // Return the minimum if (twocount < fivecount) return twocount; else return fivecount; } // Driver code let arr = [ 100, 10, 5, 25, 35, 14 ]; let n = 6; document.write(find_con_zero(arr, n)); // This code is contributed by code_hunt. </script> |
输出：
4
© 版权声明
文章版权归作者所有，未经允许请勿转载。
THE END
![关于”PostgreSQL错误：关系[表]不存在“问题的原因和解决方案-yiteyi-C++库](https://www.yiteyi.com/wp-content/themes/zibll/img/thumbnail.svg)
