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;
}