10 - summation of primes
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million. cevap : 142913828922 link #include <stdio.h> #pragma hdrstop #include <tchar.h> //--------------------------------------------------------------------------- #pragma argsused unsigned char is_number_prime(unsigned long analyses_number, unsigned long *prime_number_counter); unsigned long long sum_prime_numbers(unsigned long *prime_numbers, unsigned long *prime_number_counter, unsigned long max_number); int _tmain(int argc, _TCHAR* argv[]) { unsigned long test,analyses_number,index; unsigned long long sum_of_prime_numbers; unsigned long *prime_numbers; unsigned long i; unsigned long prime_number_counter; unsigned long max_number = 2000000; unsigned long start_point = 2; prime_numbers = (unsigned long *) calloc(max_number, sizeof(unsigned long )); memset (prime_numbers, NULL, max_number); index = 0; prime_number_counter = 0; for (analyses_number = start_point; analyses_number < max_number; ++analyses_number) { if (is_number_prime(analyses_number, &prime_number_counter) == 1) { prime_numbers[index] = analyses_number; index++; } } sum_of_prime_numbers = sum_prime_numbers(prime_numbers, &prime_number_counter, max_number); return 0; } //--------------------------------------------------------------------------- unsigned char is_number_prime(unsigned long analyses_number, unsigned long *prime_number_counter) { unsigned long counter = 0; unsigned long i; for (i = 2; i < analyses_number; i ++) { if (analyses_number%i == 0) counter++; } if (counter ! »