Find the sum of all the primes below two million. Project euler, C -
so, seems working nicely, program doesn't give me correct answer. mine 142,915,960,832, whereas should 142,913,828,922. differece 2,131,910 (if still can subtract numbers on paper haha) , have no idea did 2 millions. me?
#include <stdio.h> #include <math.h> #define below 2000000 int isaprime (int num); int main (void) { int i; float sum = 0; (i = 2; < below; i++) { if (isaprime(i) == 1) { sum = sum + i; printf ("\n%d\t%.1f", i, sum); } } getch(); return 0; } int isaprime (int num) { int i; (i = 2; <= sqrt(num); i++) { if (num % == 0) { return 0; } else { ; } } return 1; }
using float sum problem. largest integer k such integers [-k, k] exactly representable in 32-bit float 2^241; after start losing precision in integers. since sum outside range that, absurd margin, lose precision , bets off.
you need change larger type long (assuming it's 64-bits on machine). make change, , you'll right answer (as did code):
[ec2-user@ip-10-196-190-10 ~]$ cat -n euler.c 1 #include <stdio.h> 2 #include <math.h> 3 4 #define below 2000000 5 6 int isaprime (int num); 7 8 int main (void) { 9 10 int i; 11 long sum = 0; 12 13 (i = 2; < below; i++) { 14 15 if (isaprime(i) == 1) { 16 sum = sum + i; 17 } 18 } 19 printf("sum: %ld\n", sum); 20 21 return 0; 22 } 23 24 int isaprime (int num) { 25 26 int i; 27 28 (i = 2; <= sqrt(num); i++) { 29 if (num % == 0) { 30 return 0; 31 } 32 else { 33 ; 34 } 35 } 36 37 return 1; 38 } [ec2-user@ip-10-196-190-10 ~]$ gcc euler.c -lm [ec2-user@ip-10-196-190-10 ~]$ ./a.out sum: 142913828922 1: 23 explicit bits in mantissa plus 1 hidden bit.
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