给定一个数字，求其数字之和。
null
例如：
Input : n = 687 Output : 21 Input : n = 12 Output : 3
给定数字中数字和的通用算法：
- 拿到号码
- 声明一个变量来存储和，并将其设置为0
- 重复接下来的两个步骤，直到数字不为0
- 在余数“%”运算符的帮助下，将数字的最右边的数字除以10，然后将其相加。
- 在“/”运算符的帮助下，将数字除以10，以删除最右边的数字。
- 打印或返回金额
下面是获得数字总和的解决方案。 1.迭代：
C++
// C program to compute sum of digits in // number. #include <iostream> using namespace std; /* Function to get sum of digits */ class gfg { public : int getSum( int n) { int sum = 0; while (n != 0) { sum = sum + n % 10; n = n / 10; } return sum; } }; // Driver code int main() { gfg g; int n = 687; cout << g.getSum(n); return 0; } // This code is contributed by Soumik |
C
// C program to compute sum of digits in // number. #include <stdio.h> /* Function to get sum of digits */ int getSum( int n) { int sum = 0; while (n != 0) { sum = sum + n % 10; n = n / 10; } return sum; } // Driver code int main() { int n = 687; printf ( " %d " , getSum(n)); return 0; } |
JAVA
// Java program to compute // sum of digits in number. import java.io.*; class GFG { /* Function to get sum of digits */ static int getSum( int n) { int sum = 0 ; while (n != 0 ) { sum = sum + n % 10 ; n = n / 10 ; } return sum; } // Driver code public static void main(String[] args) { int n = 687 ; System.out.println(getSum(n)); } } // This code is contributed by Gitanjali |
蟒蛇3
# Python 3 program to # compute sum of digits in # number. # Function to get sum of digits def getSum(n): sum = 0 while (n ! = 0 ): sum = sum + int (n % 10 ) n = int (n / 10 ) return sum # Driver code n = 687 print (getSum(n)) |
C#
// C# program to compute // sum of digits in number. using System; class GFG { /* Function to get sum of digits */ static int getSum( int n) { int sum = 0; while (n != 0) { sum = sum + n % 10; n = n / 10; } return sum; } // Driver code public static void Main() { int n = 687; Console.Write(getSum(n)); } } // This code is contributed by Sam007 |
PHP
<?php // PHP Code to compute sum // of digits in number. // Function to get // $sum of digits function getsum( $n ) { $sum = 0; while ( $n != 0) { $sum = $sum + $n % 10; $n = $n /10; } return $sum ; } // Driver Code $n = 687; $res = getsum( $n ); echo ( "$res" ); // This code is contributed by // Smitha Dinesh Semwal. ?> |
Javascript
<script> // Javascript program to compute sum of digits in // number. /* Function to get sum of digits */ function getSum(n) { var sum = 0; while (n != 0) { sum = sum + n % 10; n = parseInt(n / 10); } return sum; } // Driver code var n = 687; document.write(getSum(n)); </script> |
输出
21
如何计算 单线？ 下面的函数有三行，而不是一行，但它计算行中的和。如果我们将指针传递给sum，它可以成为一行函数。
C++
#include <iostream> using namespace std; /* Function to get sum of digits */ class gfg { public : int getSum( int n) { int sum; /* Single line that calculates sum */ for (sum = 0; n > 0; sum += n % 10, n /= 10) ; return sum; } }; // Driver code int main() { gfg g; int n = 687; cout << g.getSum(n); return 0; } // This code is contributed by Soumik |
C
#include <stdio.h> /* Function to get sum of digits */ int getSum( int n) { int sum; /* Single line that calculates sum */ for (sum = 0; n > 0; sum += n % 10, n /= 10) ; return sum; } // Driver code int main() { int n = 687; printf ( " %d " , getSum(n)); return 0; } |
JAVA
// Java program to compute // sum of digits in number. import java.io.*; class GFG { /* Function to get sum of digits */ static int getSum( int n) { int sum; /* Single line that calculates sum */ for (sum = 0 ; n > 0 ; sum += n % 10 , n /= 10 ) ; return sum; } // Driver code public static void main(String[] args) { int n = 687 ; System.out.println(getSum(n)); } } // This code is contributed by Gitanjali |
蟒蛇3
# Function to get sum of digits def getSum(n): sum = 0 # Single line that calculates sum while (n > 0 ): sum + = int (n % 10 ) n = int (n / 10 ) return sum # Driver code n = 687 print (getSum(n)) # This code is contributed by # Smitha Dinesh Semwal |
C#
// C# program to compute // sum of digits in number. using System; class GFG { static int getSum( int n) { int sum; /* Single line that calculates sum */ for (sum = 0; n > 0; sum += n % 10, n /= 10) ; return sum; } // Driver code public static void Main() { int n = 687; Console.Write(getSum(n)); } } // This code is contributed by Sam007 |
PHP
<?php // PHP Code for Sum the // digits of a given number // Function to get sum of digits function getsum( $n ) { // Single line that calculates $sum for ( $sum = 0; $n > 0; $sum += $n % 10, $n /= 10); return $sum ; } // Driver Code $n = 687; echo (getsum( $n )); // This code is contributed by // Smitha Dinesh Semwal. ?> |
Javascript
<script> // Javascript program to compute // sum of digits in number. // Function to get sum of digits function getSum(n) { let sum; // Single line that calculates sum for (sum = 0; n > 0; sum += n % 10, n = parseInt(n / 10)) ; return sum; } // Driver code let n = 687; document.write(getSum(n)); // This code is contributed by subhammahato348 </script> |
输出
21
2.递归 感谢Ayesha提供以下递归解决方案。
算法：
1) Get the number 2) Get the remainder and pass the next remaining digits 3) Get the rightmost digit of the number with help of the remainder '%' operator by dividing it by 10 and add it to sum. Divide the number by 10 with help of '/' operator to remove the rightmost digit. 4) Check the base case with n = 0 5) Print or return the sum
C++
// C++ program to compute // sum of digits in number. #include <iostream> using namespace std; class gfg { public : int sumDigits( int no) { if (no == 0){ return 0 ; } return (no % 10) + sumDigits(no / 10) ; } }; // Driver code int main( void ) { gfg g; cout << g.sumDigits(687); return 0; } |
C
// C program to compute // sum of digits in number. #include <stdio.h> int sumDigits( int no) { if (no == 0){ return 0 ; } return (no % 10) + sumDigits(no / 10) ; } int main() { printf ( "%d" , sumDigits(687)); return 0; } |
JAVA
// Java program to compute // sum of digits in number. import java.io.*; class GFG { /* Function to get sum of digits */ static int sumDigits( int no) { if (no == 0 ){ return 0 ; } return (no % 10 ) + sumDigits(no / 10 ) ; } // Driver code public static void main(String[] args) { System.out.println(sumDigits( 687 )); } } // This code is contributed by Gitanjali |
蟒蛇3
# Python program to compute # sum of digits in number. def sumDigits(no): return 0 if no = = 0 else int (no % 10 ) + sumDigits( int (no / 10 )) # Driver code print (sumDigits( 687 )) # This code is contributed by # Smitha Dinesh Semwal |
C#
// C# program to compute // sum of digits in number. using System; class GFG { /* Function to get sum of digits */ static int sumDigits( int no) { return no == 0 ? 0 : no % 10 + sumDigits(no / 10); } // Driver code public static void Main() { Console.Write(sumDigits(687)); } } // This code is contributed by Sam007 |
PHP
<?php // PHP program to compute // sum of digits in number. function sumDigits( $no ) { return $no == 0 ? 0 : $no % 10 + sumDigits( $no / 10) ; } // Driver Code echo sumDigits(687); // This code is contributed by aj_36 ?> |
Javascript
<script> // Program to compute // sum of digits in number // Function to get sum of digits function sumDigits(no) { if (no == 0){ return 0 ; } return (no % 10) + sumDigits(parseInt(no/10)) ; } // Driver code document.write(sumDigits(687)); // This is code is contributed by simranarora5sos </script> |
输出
21
3.将输入作为字符串
当该数字的位数超过10时 19 ，我们不能将该数字作为整数，因为long int的范围不满足给定的数字。因此，将输入作为字符串，从开始到字符串长度运行一个循环，并使用该字符增加总和（在本例中为数字）
下面是上述方法的实现
C++14
// C++ implementation of the above approach #include <iostream> using namespace std; int getSum(string str) { int sum = 0; // Traversing through the string for ( int i = 0; i < str.length(); i++) { // Since ascii value of // numbers starts from 48 // so we subtract it from sum sum = sum + str[i] - 48; } return sum; } // Driver Code int main() { string st = "123456789123456789123422" ; cout << getSum(st); return 0; } |
JAVA
// Java implementation of the above approach import java.io.*; class GFG { static int getSum(String str) { int sum = 0 ; // Traversing through the string for ( int i = 0 ; i < str.length(); i++) { // Since ascii value of // numbers starts from 48 // so we subtract it from sum sum = sum + str.charAt(i) - 48 ; } return sum; } // Driver Code public static void main(String[] args) { String st = "123456789123456789123422" ; System.out.print(getSum(st)); } } // This code is contributed by Dharanendra L V. |
蟒蛇3
# Python implementation of the above approach def getSum(n): # Initializing sum to 0 sum = 0 # Traversing through string for i in n: # Converting char to int sum = sum + int (i) return sum n = "123456789123456789123422" print (getSum(n)) |
C#
// C# implementation of the above approach using System; public class GFG { static int getSum(String str) { int sum = 0; // Traversing through the string for ( int i = 0; i < str.Length; i++) { // Since ascii value of // numbers starts from 48 // so we subtract it from sum sum = sum + str[i] - 48; } return sum; } // Driver Code static public void Main() { String st = "123456789123456789123422" ; Console.Write(getSum(st)); } } // This code is contributed by Dharanendra L V. |
Javascript
<script> // Javascript implementation of the above approach function getSum(str) { let sum = 0; // Traversing through the string for (let i = 0; i < str.length; i++) { // Since ascii value of // numbers starts from 48 // so we subtract it from sum sum = sum + parseInt(str[i]); } return sum; } // Driver Code let st = "123456789123456789123422" ; document.write(getSum(st)); // This code is contributed by subhammahato348. </script> |
输出
104
4. 使用尾部递归
这个问题也可以通过尾部递归来解决。这里有一个解决方法。
1.向函数中添加另一个变量“Val”，并将其初始化为（Val=0）
2.每次调用函数时，将mod值（n%10）作为“（n%10）+val”添加到变量中，这是n中的最后一位数字。同时将变量n作为n/10传递。
3.所以在第一次通话时，它会有最后一个数字。当我们将n/10作为n传递时，它会一直跟随，直到n减少到一个位数。
4.n<10是基本情况，因此当n<10时，将n添加到变量中，因为它是最后一个数字，并返回具有数字总和的val
C++
// C++ program for the above approach #include <iostream> using namespace std; // Function to check sum // of digit using tail recursion int sum_of_digit( int n, int val) { if (n < 10) { val = val + n; return val; } return sum_of_digit(n / 10, (n % 10) + val); } // Driver code int main() { int num = 12345; int result = sum_of_digit(num, 0); cout << "Sum of digits is " << result; return 0; } // This code is contributed by subhammahato348 |
JAVA
// Java program for the above approach import java.io.*; import java.lang.*; import java.util.*; class sum_of_digits { // Function to check sum // of digit using tail recursion static int sum_of_digit( int n, int val) { if (n < 10 ) { val = val + n; return val; } return sum_of_digit(n / 10 , (n % 10 ) + val); } // Driven Program to check above public static void main(String args[]) { int num = 12345 ; int result = sum_of_digit(num, 0 ); System.out.println( "Sum of digits is " + result); } } |
蟒蛇3
# Python3 program for the above approach # Function to check sum # of digit using tail recursion def sum_of_digit(n, val): if (n < 10 ): val = val + n return val return sum_of_digit(n / / 10 , (n % 10 ) + val) # Driver code num = 12345 result = sum_of_digit(num, 0 ) print ( "Sum of digits is" , result) # This code is contributed by subhammahato348 |
C#
// C# program for the above approach using System; class GFG{ // Function to check sum // of digit using tail recursion static int sum_of_digit( int n, int val) { if (n < 10) { val = val + n; return val; } return sum_of_digit(n / 10, (n % 10) + val); } // Driver code public static void Main() { int num = 12345; int result = sum_of_digit(num, 0); Console.Write( "Sum of digits is " + result); } } // This code is contributed by subhammahato348 |
Javascript
<script> // Javascript program for the above approach // Function to check sum // of digit using tail recursion function sum_of_digit(n, val) { if (n < 10) { val = val + n; return val; } return sum_of_digit(parseInt(n / 10), (n % 10) + val); } // Driver code let num = 12345; let result = sum_of_digit(num, 0); document.write( "Sum of digits is " + result); // This code is contributed by subhammahato348 </script> |
输出
Sum of digits is 15
如果您发现上述代码/算法不正确，请写下评论，或者找到更好的方法来解决相同的问题。
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