Find digital roots of a number [closed] - c++

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This is an ACM problem in order to finding the roots of an integer number.
Here is the problem text: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=115
This is my code, but when I submit the code, I get wrong answer. In other side, I've check this code with numbers of integers and I've get the correct answer.
#include <iostream>
using namespace std;
int main() {
unsigned long long cc = 0;
cin >> cc;
while (cc != 0) {
unsigned long long sum = 0;
while (cc > 0) {
sum += cc % 10;
cc = cc / 10;
if (cc == 0 && sum > 9) { cc = sum; sum = 0; }
}
cout << sum;
cin >> cc;
cout << endl;
}
}
Can you please help me?!
Thank you.

Answer 1

The problem is that the input integer is larger than what would fit in an unsigned long long.
Therefore, you need to read the number as a string, and then calculate the digit sum from the string.
The following code will work:
#include <iostream>
#include <string>
using namespace std;
int main()
{
string inStr;
while(cin >> inStr && inStr != "0")
{
unsigned long long cc = 0;
for(string::const_iterator it = inStr.begin(); it!=inStr.end(); ++it)
{
cc += *it - '0';
}
unsigned long long sum = 0;
do
{
while (cc)
{
sum += cc % 10;
cc = cc / 10;
}
cc = sum;
sum = 0;
}while(cc > 9);
cout << cc << endl;
}
return 0;
}

Answer 2

I wonder why anyone didn't post this yet... :P
the function returns the answer :)
int Digital_root(int a) {
return a%9==0 ? 9:a%9;
}

Answer 3

Probably the problem is in that number might contain more than 2 digits and in this case such a modification is necessary:
int main() {
unsigned long long cc = 0;
cin >> cc;
unsigned long long sum = 0;
while (cc > 0) {
sum += cc % 10;
cc = cc / 10;
if ( sum > 9) { cc = sum; sum = 0; }
^
// cc == 0 will fail
}
cout << sum;
}

Answer 4

It's not perfect code, but can work
int a = 0;
int b = 0;
while (true)
{
cout << endl << "a: ";
cin >> a;
if (!a) break;
do
{
while (a)
{
b += a%10;
a /= 10;
}
a = b;
b = 0;
}
while (a > 9);
cout << endl<< "root: " << a;

Answer 5

The task really asks for the remainder under division by 9.
Reason: Since 10 mod 9 == 1 and thus also 10^k mod 9 == 1, the sum of decimal digits has the same remainder under division by 9 as the number itself. Repeated sums of digits do not change the remainder, so the decimal digital root of some n is the same as n mod 9 or computing the digital sum of n modulo 9.
Reducing the code of riklund to this basic task gives
#include <iostream>
#include <string>
using namespace std;
int main() {
string inStr;
while(cin >> inStr && inStr != "0") {
unsigned int cc = 0; // need only 5 bit for cc in this computation
for(string::const_iterator it = inStr.begin(); it!=inStr.end(); ++it) {
cc += *it - '0';
cc %=9;
}
cout << cc << endl;
}
return 0;
}

Answer 6

#include<iostream>
using namespace std;
int main()
{
for (int i = 0; ; i++)
{
unsigned long int x,sum=0;
cin >> x;
if (x == 0)
break;
if (x <= 9)
{
sum = x;
goto z;
}
while (x > 9)
{
while (x != 0)
{
sum = sum + (x%10);
x = x / 10;
}
if (sum > 9)
{
x = sum;
sum = 0;
}
}
z:
cout << sum <<"\n";
}
}

Answer 7

//digital roots.cpp~KAUSHIK
#include<iostream>
using namespace std;
int sum(int n)
{
int sum=0,r;
for (;n>0;)
{
r=n%10;
sum=sum+r;
n=n/10;
}
return sum;
}
int main()
{
int n;
cout<<"enter any number"<<endl;
cin>>n;
int a=n;
n=sum(n);
if((n/10)!=0)
{
n=sum(n);
cout<<"the digital root of "<<a<<" is"<<n;
}
else cout<<"the digital root of "<<a<<" is"<<n;
return 0;
}

Answer 8

it works for small integers
#include <iostream>
using namespace std;
int main()
{
unsigned long long input;
while (true)
{
cin >> input;
if (input == 0) break;
input = input - (9 * ((input - 1) / 9));
cout << input << endl;
}
return 0;
}

Answer 9

it works simple copy that in main(), sorry for my english.
int a = 39;
int b = 0;
do
{
while (a)
{
b += a%10;
a /= 10;
}
a = b;
b = 0;
}
while (a > 9);