ZJU 1115 解题报告

Digital Roots

---

Time limit: 1 Seconds   Memory limit: 32768K  
Total Submit: 12051   Accepted Submit: 2591  

---

Background

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.

Example

Input

24
39
0

Output

6
3

---

Problem Source: Greater New York 2000

题目地址：http://acm.zju.edu.cn/show_problem.php?pid=1115

本题大意是对于给定的一个正数，对其各位数字相加。如果结果为个位数，则输出，反之对结果的各位数字相加，直至得到一个个位数的结果。

本题的关键在于不知道输入的数有多少位，所以不能用int型，只能用数组存储输入的数。对于数组要开多大也不清楚，我开到511后还是提示Runtime Error SIGSEGV，直到1023后才AC。参考代码如下：

#include  < stdio.h >

int  main( int  argc,  char   * argv[])
... {
        int i, n;
    char digit[1023];

    while (1) ...{
        gets(digit);
        for (n = 0, i = 0; digit[i] != '/0'; i ++) n += digit[i] - 48;
        if (!n) break;
        while (n > 9) ...{
            for (i = 0; n; n /= 10) i += n % 10;
            n = i;
        }
        printf("%d ", n);
    } 

    return 0;
}

 

附：SIGSEGV in FAQ (http://acm.zju.edu.cn/faq.php)

 

---

 

Q:What does SIGSEGV in Runtime Error stand for?
A:The following messages will not be shown to you in contest. Here we just provide some tips:
SIGSEGV --- Segment Fault. The possible cases of your encountering this error are:

• 1.buffer overflow --- usually caused by a pointer reference out of range.
  
• 2.stack overflow --- please keep in mind that the default stack size is 8192K.
  
• 3.illegal file access --- file operations are forbidden on our judge system.
  

SIGFPE  --- Divided by 0
SIGBUS  --- Hardware Error. //please contact us
SIGABRT --- Programme aborted before it should be finished.
man 7 signal under Linux for more information

---