杭电1003

Problem Description

The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits are summed and the process is repeated. This is continued as long as necessary to obtain a single digit.

For example, consider the positive integer 24. Adding the 2 and the 4 yields a value of 6. Since 6 is a single digit, 6 is the digital root of 24. Now consider the positive integer 39. Adding the 3 and the 9 yields 12. Since 12 is not a single digit, the process must be repeated. Adding the 1 and the 2 yeilds 3, a single digit and also the digital root of 39.

 

Input

The input file will contain a list of positive integers, one per line. The end of the input will be indicated by an integer value of zero.

 

Output

For each integer in the input, output its digital root on a separate line of the outpu

 

Sample Input

  
  

   
   24
39
0
  
  

 

Sample Output

  
  

   
   6
3
  
  
  
  

   
   开始时，我的错误代码：
  
  
  
  

   
   #include<stdio.h>
int main(){

   
   	int n;

   
   	while(scanf("%d",&n)!=EOF&&n!=0){

   
   		if(n<9){

   
   			printf("%d\n",n);

   
   		}

   
   		else{

   
   			if(n%9==0)

   
   			   printf("9\n");

   
   			else

   
   			   printf("%d\n",n%9);//可以分为两种情况考虑

   
   		}

   
   	}

   
   	return 0;

}没有考虑n为大数，
   
   Ø
   
   输入正整数可能是大数，需要用字符串运算。
   
   

    
    Ø
    
    将正整数各个位相加，如果结果为个位数则结束，否则继续相加，直至为个位数为止。
   
   
   
   

    
    Ø
    
     可以利用模
    
    9
    
    运算。
   
   
   
   

    
    正确代码：
   
   
   
   

    
    #include<stdio.h>
int main(){

    
    	char ch;

    
    	while(1){//代表循环一直进行，除非有break跳出循环

    
    		int sum=0;

    
    		while(scanf("%c",&ch)&&ch!='\n')

    
    		sum+=ch-'0';//将字符型变量化为整形

    
    		if(sum==0) 输入0时，循环结束

    
    		  break;

    
    		if(sum%9==0)

    
    		  sum=9;

    
    		else

    
    		  sum=sum%9;

    
    		printf("%d\n",sum);
    }
        return 0;
}

   
   
   
   

    
    运行结果：