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Diffstat (limited to '09-Special-Pythagorean-Triplet/pyth.c')
-rw-r--r-- | 09-Special-Pythagorean-Triplet/pyth.c | 24 |
1 files changed, 24 insertions, 0 deletions
diff --git a/09-Special-Pythagorean-Triplet/pyth.c b/09-Special-Pythagorean-Triplet/pyth.c new file mode 100644 index 0000000..86a12b0 --- /dev/null +++ b/09-Special-Pythagorean-Triplet/pyth.c @@ -0,0 +1,24 @@ +#include <stdio.h> +#include <stdlib.h> +#include <math.h> + +/* Actually found a neat math solution to make this one quicker + * since (a + b + c) = 1000 + * and a^2 + b^2 = c^2 + * + * We can say: + * c = 1000 - (a + b) + * c^2 = (1000 - (a + b))^2 + * and + * a^2 + b^2 = (1000 - (a + b))^2 + * So you need to find an a and b that satisfy this + * We know a and b are less than 500 since c must be larger +*/ + +int main(){ + for(long a = 1; a < 500; a++) + for(long b = 1; b < 500; b++) + if(pow(a, 2) + pow(b, 2) == pow((1000 - (a + b)), 2)) + return printf("%d\n", a * b * (long) sqrt(pow(a, 2) + pow(b, 2))); + return 1; +} |