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Henrik Hautakoski 2010-04-06 06:48:23 +02:00
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/*
* http://projecteuler.net
*
* Projecteuler - Problem 9
* ------------------------
* 2010-04-06 Henrik Hautakoski
*
* the math stuff here (simple algebra)
*
* 1000 = a + b + c
* 10^6 = (a + b + c)^2
* 10^6 = a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2
* 10^6 = a^2 + b^2 + c^2 + 2(ab + ac + bc)
* 10^6 = 2a^2 + 2b^2 + 2(ab + ac + bc) (substitution: a^2 + b^2 = c^2, pythagorean theorem)
* 10^6/2 = a^2 + b^2 + ab + ac + bc
* 10^6/2 + ab = a^2 + b^2 + 2ab + ac + bc
* 10^6/2 + ab = (a + b)(a + b) + (a + b)c
* 10^6/2 + ab = (a + b + c)(a + b)
* 10^6/2 + ab = 1000a + 1000b (substitution: a + b + c = 1000)
* 10^6/2 = 1000a + 1000b - ab
* 10^6/2 = a(1000 - b) + 1000b
*
* Equation to use: a = (10^6/2 - 1000b)/(1000 - b)
*/
#include <stdio.h>
int main() {
int a, b;
double tmp;
for(b=1; b < 1000; b++) {
a = tmp = (5e5 - 1000*b)/(1000 - b);
if (tmp - a >= 0.0 && (tmp - a) < .00000001) {
printf("%i\n", a*b*(1000 - (a+b)));
break;
}
}
return 0;
}