Created
May 3, 2011 17:24
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Simple mathematical approach for summing the digits in a number
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#include <stdio.h> | |
#include <math.h> | |
/* | |
The algorithm is: | |
d_i = (c mod 10^(i+1)) / 10^i | |
S = sum_{i=0}^{floor(log_10(c))+1} d_i = sum_{i=0}^{floor(log_10(c))+1} (c mod 10^(i+1)) / 10^i | |
Where c is the number in question, d_i is the digit at position i counted from the least significant digit | |
and S is the sum of the digits in the number. | |
*/ | |
int main(void) | |
{ | |
int c = 1000; | |
printf("The number is %d.\n", c); | |
int s = 0; | |
/* | |
// Special case for 3 digits: | |
s += (c % 10) / 1; | |
s += (c % 100) / 10; | |
s += (c % 1000) / 100; | |
*/ | |
printf("%f\n", log(c)); | |
// General case for n digits | |
int i; | |
for (i = 0; i < (int)(log(c) / log(10)) + 1; i++) { | |
s += (c % (int)pow(10, i + 1)) / (int)pow(10, i); | |
} | |
printf("The sum of the digits is %d.\n", s); | |
return 0; | |
} | |
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