HDU 1013 Digital Roots

Digital Roots

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

Total Submission(s): 62745    Accepted Submission(s): 19486


[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
For each integer in the input, output its digital root on a separate line of the output.

 

[align=left]Sample Input[/align]

24
39
0

 

[align=left]Sample Output[/align]

6
3

分析：

一个整数，把所有位都加起来，得出的数如果小于10就输出，不小于继续把所有位都加起来。

该题目是个数论题，但是由于没有接触过，所以没用公式写，直接模拟出来的。

注意几个问题就可以了：输入的数字会非常大，要用字符存储，长度为1000左右，判断时注意判断条件是‘>=10’

代码如下：

<span style="font-size:18px;">#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int n,temp;
int i,len;
char str[1005];
while(scanf("%s",&str),str[0]!='0')
{
len=strlen(str);
temp=0;
for(i=0;i<len;i++)
temp+=str[i]-'0';
while(temp>=10)
{
memset(str,0,sizeof(str));
sprintf(str,"%d",temp);
len=strlen(str);
temp=0;
for(i=0;i<len;i++)
temp+=str[i]-'0';
}

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

return 0;
}</span>