HDU 1013 Digital Roots

Digital Roots

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29353 Accepted Submission(s): 8974


[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

[align=left]Source[/align]
Greater New York 2000

//常规做法

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int del(int a)
{
int b=0;
while(a>0)
{
b+=a%10;
a/=10;
}
return b;
}
int main()
{
char s[1003];
int i;
int a;
while(scanf("%s",s),s[0]!='0')
{   a=0;
for(i=0;s[i]!='\0';i++)
a+=s[i]-'0';
while(a>9)
{
a=del(a);
}
printf("%d\n",a);
}
return 0;
}

//快捷算法
//算法思想：
//一个数 模9等于各位数字和模9，例如 33%9=6%9；
//证明： a1a2a3...an%9=((a1*10^n-1)%9+(a2*10^n-2)%9...)%9
//右边 a1*(9999..9+1)%9=a1%9，以此类推、、、、、
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
char s[1003];
int i;
int a;
while(scanf("%s",s),s[0]!='0')
{   a=0;
for(i=0;s[i]!='\0';i++)
a=a*10+s[i]-'0',a=a%9;
if(a==0) a=9;
printf("%d\n",a);
}
return 0;
}