hi
if i have a list of numbers , say 2345678
and want to add up the odd position digits
ie 2+4+6+8
how can i do this ?
i tried it with nums[20]
and using atoi to convert the type
but it seems i can't do things like
sum=atoi(nums[0]) + atoi(nums[1]);
can someone plz help me~~~~
Try this:
PHP Code:char nums[]="2345678"
int i;
int add=0;
for(i=0; i < strlen(nums); i + 2)
{
add = add + (atoi(nums[i]) ); /// OR add += (atoi(nums[i]) )
}
printf("%i", add);
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
try changing :-
for(i=0; i < strlen(nums); i + 2)
to
for(i=0; i < strlen(nums); i += 2)
Free the weed!! Class B to class C is not good enough!!
And the FAQ is here :- http://faq.cprogramming.com/cgi-bin/smartfaq.cgi
this will do it for any number
Code:#include <stdlib.h> #include <string.h> #define MAX_SIZE 15 int function(int num) { int ret = 0; char number[MAX_SIZE]; memset(number,'\0',MAX_SIZE); _itoa(num,number,10); for(int i = 0;number[i] != '\0';i++) { if((i % 2 != 0) || i == 1) ret += i; } return ret; }
ADVISORY: This users posts are rated CP-MA, for Mature Audiences only.
Does it have to be a string
int num[7] ={2,3,4,5,6,7,8};
then
for(i =0; i < 7;i++)
if(i % 2 == 0)
total += num[i];
should do the trick
hoping to be certified (programming in c)
here's the news - I'm officially certified.
A third way.
Code:#include <stdlib.h> #include <stdio.h> int add_odd_digits(int number); int main() { int num = 12345; int sum; sum = add_odd_digits(num); printf("sum of odd digits:%d\n",sum); return 0; } int add_odd_digits(int number) { int digit, odd_sum; int sum1 = 0; int sum2 = 0; while (number > 0) { digit = number % 10; number /= 10; sum1 += digit; if (number <=0) return sum1; digit = number % 10; number /= 10; sum2 += digit; if (number <=0) return sum2; } }
