int sumDigits(int n){
int rc=0;
if(n > 0){
rc=n%10 + sumDigits(n/10);
}
return rc;
}
/**
* Given a non-negative int n, return the sum of its digits recursively (no loops).
*
* @param n The number to sum the digits of
* @return The sum of its digits
*
* @note Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6),
* while divide (/) by 10 removes the rightmost digit (126 / 10 is 12).
*
* Example:
* sumDigits(126) == 9
* sumDigits(49) == 13
* sumDigits(12) == 3
*/
int RecursionExercises::sumDigits(int n)
{
if (n<=0)
return 0;
return n%10 + sumDigits((n/10));
}
// Given a non-negative int n, return the sum of its digits recursively (no loops).
int sumDigits( int n )
{
if( n <= 0 )
return 0;
return ( n % 10 ) + sumDigits( n / 10 );
}
int sumDigits(int n)
{
/*if(n!=0)
{
int m;
m=n%10;
n=n/10;
m=(sumDigits(n)+m);
return m;
}
else
{ int sum=0;
return sum;; // Replace this with your code
}
*/
if(n==0) return 0;
else
{
int m=n%10;
n=n/10;
return m+sumDigits(n);
}
}
/**
* Given a non-negative int n, return the sum of its digits recursively (no loops).
*
* @param n The number to sum the digits of
* @return The sum of its digits
*
* @note Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6),
* while divide (/) by 10 removes the rightmost digit (126 / 10 is 12).
*
* Example:
* sumDigits(126) == 9
* sumDigits(49) == 13
* sumDigits(12) == 3
*/
int RecursionExercises::sumDigits(int n)
{
if (n==0) return 0;
int temp = n;
temp = n%10;
temp = temp + sumDigits(n/10);
return temp;
}
/**
* Given a non-negative int n, return the sum of its digits recursively (no loops).
*
* @param n The number to sum the digits of
* @return The sum of its digits
*
* @note Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6),
* while divide (/) by 10 removes the rightmost digit (126 / 10 is 12).
*
* Example:
* sumDigits(126) == 9
* sumDigits(49) == 13
* sumDigits(12) == 3
*/
int RecursionExercises::sumDigits(int n)
{
if(n < 10){
return n;
}
return sumDigits(n/10) + n % 10;
}
int main(void)
{
int n;
printf("Enter number to sum up: ");
scanf("%d", &n);
printf("%d \n", sumDigits(n));
getchar();
return 0;
}
int main()
{
int num = 3748;
printf("Ingresa un numero: ");
scanf("%d", &num);
printf("El numero: %d tiene %d digitos.\n", num, countDigits(num));
printf("El numero: %d sus numeros suman: %d\n", num, sumDigits(num));
return 0;
}
int sf(int n) {
// first check if value is in lookup table
// int lookup = getLookupValue(table_f, n);
// if( lookup != -1) {
// return lookup; }
// else {
int fn = f(n);
// AddToTable(table_f,n,fn);
return sumDigits( fn );
// }
}
bool isHappy(int n) {
if (n<0) return isHappy(-n);
set<int> myset;
while (n!=1) {
if (myset.find(n)!=myset.end()) return false;
else {
myset.insert(n);
n = sumDigits(n);
}
}
return true;
}
/**
* Given a non-negative int n, return the sum of its digits recursively (no loops).
*
* @param n The number to sum the digits of
* @return The sum of its digits
*
* @note Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6),
* while divide (/) by 10 removes the rightmost digit (126 / 10 is 12).
*
* Example:
* sumDigits(126) == 9
* sumDigits(49) == 13
* sumDigits(12) == 3
*/
int RecursionExercises::sumDigits(int n)
{
if(n==0){
return 0; //base case!
}
else
return sumDigits(n/10)+ n%10;
//n%10 gives the right most digit and n/10 cuts if off so add the two and were good to go!
}
int main()
{
// get input from user and save it to variable
printf("Enter some number\n");
long long input = GetLongLong();
unsigned long sum = sumDigits(input);
// display result
printf("Sum is %lu\n", sum);
return 0;
}
int main(){
uint limit = 30;
vector<ullint> v;
for(ullint i = 0; i < 70; i++){
for(ullint j = 0; j < 30; j++){
ullint a = pow(i, j);
ullint s = sumDigits(intToString(a));
if(s == i)
v.push_back(a);
}
}
removeDuplicates(v);
cout << v[limit + 9] << endl;
return 0;
}
int main(int argc, const char * argv[])
{
int sizeofarray = 2*numbertofactorial;
int* product;
product = new int[sizeofarray];
//initialize product array to 0s
for (int i = 0; i < sizeofarray; i++) {
product[i] = 0;
}
product[0] = 1;
for (int multiplier = 2; multiplier <= numbertofactorial; multiplier++) {
// printf("%d! = ", multiplier-1);
// printCurrentProduct(product, sizeofarray);
// ^uncomment above to see each factorial in succession^
//find the index of the zero before the first nonzero
int firstzeroinarray = sizeofarray;
while (product[firstzeroinarray-1] == 0) firstzeroinarray--;
int carry = 0;
for (int index = 0; index < firstzeroinarray; index++) {
product[index] = product[index]*multiplier + carry;
if (product[index]/10 != 0) {
carry = product[index]/10;
product[index] %= 10;
}
else carry = 0;
}
product[firstzeroinarray] = carry;
product = fixCarryIfNecessary(product, firstzeroinarray);
}
printf("\n%d! = ", numbertofactorial);
printCurrentProduct(product, sizeofarray);
int sum = sumDigits(product, sizeofarray);
printf("The sum of the digits of %d! is %d.\n", numbertofactorial, sum);
return 0;
}
int sumDigits(int no)
{
return no == 0 ? 0 : no%10 + sumDigits(no/10) ;
}
/**
* Given a non-negative int n, return the sum of its digits recursively (no loops).
*
* @param n The number to sum the digits of
* @return The sum of its digits
*
* @note Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6),
* while divide (/) by 10 removes the rightmost digit (126 / 10 is 12).
*
* Example:
* sumDigits(126) == 9
* sumDigits(49) == 13
* sumDigits(12) == 3
*/
int RecursionExercises::sumDigits(int n)
{ if(n/10 == 0)
return n%10;
else
return (sumDigits(n/10)+n%10);
}
int sg( int i ) {
sumDigits ( g ( i ) );
}
// sum of digits
int sumDigits(int x) {
if (x < 10) return x;
return sumDigits(x/10) + sumDigits(x%10);
}