[解题报告]HDU 1013 Digital Roots

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35288    Accepted Submission(s): 10794

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

24 39 0

 

Sample Output

6 3

 

Source

Greater New York 2000

 

 

 

 

 

 

 

 

 

如果你一开始把所有的值设置为int型，恭喜你，你会得到一个WA

接着你大概会改成unsigned型，恭喜你，你会得到一个超时

比如这样：

#include<stdio.h>
unsigned fun1(unsigned i)
{
    unsigned j,temp;
    for(j=1;j<i;j*=10)
    {
        if(i>j&&i<=j*10)
        {
            temp=j;
        }
    }
    unsigned sum=0;
    while(i)
    {
        sum+=i%10;
        i/=10;
    }
    return sum;
}
int main()
{
    unsigned n;
    while(scanf("%u",&n)!=EOF&&n)
    {
        while(n>=10)
        {
            n=fun1(n);
        }
        printf("%u\n",n);
    }
    return 0;
}

然后你终于恍然大悟，用字符串！！

 

#include <stdio.h>
int main()
{
char s[1000];
int i,sum;
while(scanf("%s",s)!=EOF)
{
      if(s[0]=='0')
       {
         break;
        }
      sum=0;
   for(i=0;s[i]!='\0';i++)
    {
      sum+=s[i]-'0';
    if(sum>9)
    {
     sum=sum%10+sum/10;
    }
    }
    printf("%d\n",sum);
    }
    return 0;
}