杭电 1013 Digital Roots （字符串数组）

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 48103    Accepted Submission(s): 14939

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 output.

 

Sample Input

24390

 

Sample Output

63

 

Source

Greater New York 2000

题意：

给定一个数字，当位数大于一的时候，每一位的数字相加，最后得到的数字是一位时，

输出，否则继续执行上述操作，知道得到一位数。

难点： 

这道题有个小陷阱 需要特别注意 就是要考虑到大数的问题 通过一般的类型肯定要溢出

思路：

可以利用 字符型数组 进行运算 至于判断是否是两或多位数的时候 可以通过递归来判定

代码如下：

<span style="font-size:14px;">#include<stdio.h>#include<string.h>#include<stdlib.h>int dr(int n){    int t=0;    while(n)    {            t+=n%10;            n/=10;            }//这个式子在不确定位数的时候也能将一个数的各个位数分离开来            if(t/10)            dr(t);            else            return t;    }int main(){    char a[1010];    while(~scanf("%s",a))    {                         int sum=0,i;                         if(a[0]=='0')                         break;                         else                         {                             for(i=0;i<strlen(a);i++)                             sum+=a[i]-'0';                             printf("%d\n",dr(sum));                             }                         }    return 0;    }</span>