Digital Roots

Digital Roots

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 65092 Accepted Submission(s): 20292


[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

本题我刚开始没有考虑输入的整数无穷大情况，就固定了输入数据类型，后来改为string，但是求得和还是可能无穷大，这一点没考虑但是程序居然通过了，汗

#include<iostream>
#include<string>

using namespace std;

int getRootInt(int value)
{
int sum=0;
int tempValue=value;
while(tempValue!=0)
{
sum+=tempValue%10;
tempValue=tempValue/10;
}
if(sum>9)
{
sum=getRootInt(sum);
}
return sum;
}

int getRoot(string value)
{
int sum=0;
for(int i=0;i<value.length();i++)
{
sum+=value[i]-'0';
}
return getRootInt(sum);
}

int main()
{
string value="";//不能定义固定类型可能无穷大
while(cin>>value)
{
if(value=="0")break;
int sum=0;
sum=getRoot(value);
cout<<sum<<endl;
}
return 0;
}