Project Euler Problem 73
71とほぼ一緒。
上限と下限が決め打ちされている分71より簡単。
natural f73(){ natural lowd = 3; natural lown = 1; natural highd = 2; natural highn = 1; natural counter = 0; for(natural d = 12000; d > 1 ; --d){ for(natural n = (highn * d / highd) + 1; n > lown * d / lowd - 1; --n){ if(gcd(n, d) != 1){ continue; } if(highn * d > n * highd){ if(lown * d < n * lowd){ //std::cout << n << " " << d <<std::endl; ++counter; }else { break; } } else { continue; } } } return counter; }