给定一个数字,任务是找出这个数字是否可以被9整除。输入的数字可能很大,即使使用long int,也可能无法存储。
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例如:
Input : n = 69354Output : YesInput : n = 234567876799333Output : NoInput : n = 3635883959606670431112222Output : No
由于输入的数字可能非常大,我们不能使用n%9来检查一个数字是否可以被9整除,尤其是在C/C++等语言中。这个想法基于以下事实。
A number is divisible by 9 if sum of its digits is divisible by 9.
插图:
For example n = 9432Sum of digits = 9 + 4 + 3 + 2 = 18Since sum is divisible by 9,answer is Yes.
这是怎么回事?
Let us consider 1332, we can write it as1332 = 1*1000 + 3*100 + 3*10 + 2The proof is based on below observation:Remainder of 10i divided by 9 is 1So powers of 10 only results in remainder 1 when divided by 9.Remainder of "1*1000 + 3*100 + 3*10 + 2"divided by 9 can be written as : 1*1 + 3*1 + 3*1 + 2 = 9The above expression is basically sum ofall digits.Since 9 is divisible by 9, answer is yes.
下面是上述想法的实现。
C++
// C++ program to find if a number is divisible by // 9 or not #include<bits/stdc++.h> using namespace std; // Function to find that number divisible by 9 or not int check(string str) { // Compute sum of digits int n = str.length(); int digitSum = 0; for ( int i=0; i<n; i++) digitSum += (str[i]- '0' ); // Check if sum of digits is divisible by 9. return (digitSum % 9 == 0); } // Driver code int main() { string str = "99333" ; check(str)? cout << "Yes" : cout << "No " ; return 0; } |
JAVA
// Java program to find if a number is // divisible by 9 or not class IsDivisible { // Function to find that number // is divisible by 9 or not static boolean check(String str) { // Compute sum of digits int n = str.length(); int digitSum = 0 ; for ( int i= 0 ; i<n; i++) digitSum += (str.charAt(i)- '0' ); // Check if sum of digits is divisible by 9. return (digitSum % 9 == 0 ); } // main function public static void main (String[] args) { String str = "99333" ; if (check(str)) System.out.println( "Yes" ); else System.out.println( "No" ); } } |
Python3
# Python 3 program to # find if a number is # divisible by # 9 or not # Function to find that # number divisible by 9 # or not def check(st) : # Compute sum of digits n = len (st) digitSum = 0 for i in range ( 0 ,n) : digitSum = digitSum + ( int )(st[i]) # Check if sum of digits # is divisible by 9. return (digitSum % 9 = = 0 ) # Driver code st = "99333" if (check(st)) : print ( "Yes" ) else : print ( "No" ) # This code is contributed by Nikita Tiwari. |
C#
// C# program to find if a number is // divisible by 9 or not. using System; class GFG { // Function to find that number // is divisible by 9 or not static bool check(String str) { // Compute sum of digits int n = str.Length; int digitSum = 0; for ( int i = 0; i < n; i++) digitSum += (str[i] - '0' ); // Check if sum of digits is // divisible by 9. return (digitSum % 9 == 0); } // main function public static void Main () { String str = "99333" ; if (check(str)) Console.Write( "Yes" ); else Console.Write( "No" ); } } // This code is Contributed by // nitin mittal. |
PHP
<?php // PHP program to find if a number // is divisible by 9 or not // Function to find that // number divisible by 9 or not function check( $str ) { // Compute sum of digits $n = strlen ( $str ); $digitSum = 0; for ( $i = 0; $i < $n ; $i ++) $digitSum += ( $str [ $i ] - '0' ); // Check if sum of digits // is divisible by 9. return ( $digitSum % 9 == 0); } // Driver code $str = "99333" ; $x = check( $str ) ? "Yes" : "No " ; echo ( $x ); // This code is contributed by Ajit. ?> |
Javascript
<script> // Javascript program to find if a number // is divisible by 9 or not // Function to find that // number divisible by 9 or not function check(str) { // Compute sum of digits let n = str.length; let digitSum = 0; for (let i = 0; i < n; i++) digitSum += (str[i] - '0' ); // Check if sum of digits // is divisible by 9. return (digitSum % 9 == 0); } // Driver code let str = "99333" ; let x = check(str) ? "Yes" : "No " ; document.write(x); // This code is contributed by _saurabh_jaiswal. </script> |
输出:
Yes
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