给定一个数字n,任务是在n除以11时求余数。数字的输入可能非常大。 例如:
null
Input : str = 13589234356546756Output : 6Input : str = 3435346456547566345436457867978Output : 4
由于给定的数字可能非常大,我们不能使用n%11。需要使用以下步骤来查找余数:
1. Store number in string.2. Count length of number string.3. Convert string character one by one into digit and check if it's less than11. Then continue for next character otherwise take remainder and use remainder for next number.4. We get remainder.Ex. str = "1345" len = 4 rem = 3
C++
// CPP implementation to find remainder // when a large number is divided by 11 #include <bits/stdc++.h> using namespace std; // Function to return remainder int remainder(string str) { // len is variable to store the // length of number string. int len = str.length(); int num, rem = 0; // loop that find remainder for ( int i = 0; i < len; i++) { num = rem * 10 + (str[i] - '0' ); rem = num % 11; } return rem; } // Driver code int main() { string str = "3435346456547566345436457867978" ; cout << remainder(str); return 0; } |
JAVA
// JAVA implementation to find remainder // when a large number is divided by 11 import java.io.*; class GFG{ // Function to return remainder static int remainder(String str) { // len is variable to store the // length of number string. int len = str.length(); int num, rem = 0 ; // loop that find remainder for ( int i = 0 ; i < len; i++) { num = rem * 10 + (str.charAt(i) - '0' ); rem = num % 11 ; } return rem; } // Driver code public static void main(String args[]) { String str = "3435346456547566345436457867978" ; System.out.println(remainder(str)); } } /*This code is contributed by Nikita Tiwari.*/ |
Python3
# Python 3 implementation to find remainder # when a large number is divided by 11 # Function to return remainder def remainder(st) : # len is variable to store the # length of number string. ln = len (st) rem = 0 # loop that find remainder for i in range ( 0 , ln) : num = rem * 10 + ( int )(st[i]) rem = num % 11 return rem # Driver code st = "3435346456547566345436457867978" print (remainder(st)) # This code is contributed by Nikita Tiwari. |
C#
// C# implementation to find remainder // when a large number is divided by 11 using System; class GFG { // Function to return remainder static int remainder( string str) { // len is variable to store the // length of number string. int len = str.Length; int num, rem = 0; // loop that find remainder for ( int i = 0; i < len; i++) { num = rem * 10 + (str[i] - '0' ); rem = num % 11; } return rem; } // Driver code public static void Main() { string str = "3435346456547566345436457867978" ; Console.WriteLine(remainder(str)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP implementation to find remainder // when a large number is divided by 11 // Function to return remainder function remainder( $str ) { // len is variable to store the // length of number string. $len = strlen ( $str ); $num ; $rem = 0; // loop that find remainder for ( $i = 0; $i < $len ; $i ++) { $num = $rem * 10 + ( $str [ $i ] - '0' ); $rem = $num % 11; } return $rem ; } // Driver code $str = "3435346456547566345436457867978" ; echo (remainder( $str )); // This code is contributed by Ajit. ?> |
Javascript
<script> // Javascript implementation to find remainder // when a large number is divided by 11 // Function to return remainder function remainder(str) { // len is variable to store the // length of number string. let len = str.length; let num; let rem = 0; // loop that find remainder for (let i = 0; i < len; i++) { num = rem * 10 + (str[i] - '0' ); rem = num % 11; } return rem; } // Driver code let str = "3435346456547566345436457867978" ; document.write(remainder(str)); // This code is contributed by _saurabh_jaiswal. </script> |
输出:
4
时间复杂性: O(L)式中,L是弦的长度
辅助空间: O(L)
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