Several C language implementations

root

Title Description
The number root can be obtained by adding up the numbers on each bit of a number . If the number you get is a single digit , Then this number is the number . If the result is two digits or more , Then add up these numbers . Go on like this , Until you get a single digit .

such as , about 24 Come on , hold 2 and 4 Add up to get 6, because 6 It's one digit. , therefore 6 yes 24 The number of roots . Another example 39, hold 3 and 9 Add up to get 12, because 12 Not a single digit , So we have to 1 and 2 Add up , Finally get 3, This is a single digit , therefore 3 yes 39 The number of roots .

Input ：
A positive integer ( Less than 10 Of 10000 Power ).

Output ：
A number , That is, input several roots of numbers .

The sample input ：
24

Sample output ：

6

Because the number is too big , So you can use character arrays to write .
The code is as follows ：

#include<stdio.h>
int main()
{
	char num[1001];
	int s,i;
	scanf("%s",num);
	while(num[1]!='\0')        // Jump out of the loop when there is only one digit left in the number , Output number 
	{
		s=0;
		i=0;
		while(num[i]!='\0')     
		{
			s=s+(num[i]-48);   // Convert characters to numbers and add them up 
			i++;	
		}
		i=0;
		if(s==0)
			break;
		while(s!=0)
		{
			num[i++]=(s%10)+48;     // Convert the added numbers into characters and store them in the array again 
			s=s/10;
		}
		num[i]='\0';
	}
	printf("%d\n",num[0]-48);           // Convert the final result into digital output 
	return 0;
	
}

Main idea ：
Convert the characters stored in the character array into numbers and add the numbers on each bit , Then it is converted into character array storage , Loop to the array with only one significant digit left , Convert it to digital output .

because 10 Of 10000 Power 9 Addition is also int Within type , So you don't have to use an array to store , After that, you can continue to find the remainder, round it and add it
Optimize ：

#include<stdio.h>
int main()
{
    
	char a[1010];
	int n=0;
	scanf("%s",a);
	for(int i=0; a[i]!='\0'; i++)
		n +=  a[i]-'0';
	while(n>=10)
	{
    
		int m=0;
		while(n!=0)
		{
    
			m += n%10;
			n = n/10;
		}	
		n=m;
	}
	printf("%d\n",n);
	return 0;
}