HDU 4611

对于任意a,b，只要最多计算一遍循环节就行了。

对于这样的题目，明了的程序结构比解法更重要。

 

#include <cstdio>
#include <cstring>

using std::memset;

long long t, n, a, b;

long long max(long long a, long long b)
{
	return a < b? b : a;
}

long long min(long long a, long long b)
{
	return a < b? a : b;
}

long long abs(long long a)
{
	return a > 0? a : -a;
}

long long gcd(long long a, long long b)
{
	return b?gcd(b, a%b): a;
}

long long count(long long l,long long a, long long b)
{
	long long ans = 0;
	long long sa = 0, sb = 0;

	while(sa < l && sb < l)
	{
		long long from = max(sa, sb);
		long long to = min(sa+a, sb+b);
		
		if(to >= l)
		{
			break;
		}

		ans += (to - from) * abs( from % a - from % b);
		if(sa + a < sb + b)
		{
			sa = to;
		}
		else
		{
			sb = to;
		}
	}
	long long from = max(sa, sb);
	ans += (l - from) * abs(from%a - from%b);
	return ans;
}

int main()
{
	scanf("%I64d", &t);
	while(t--)
	{
		scanf("%I64d%I64d%I64d", &n, &a, &b);
		long long ans = 0;
		long long lcm = a*b/gcd(a, b);
		if( n > lcm)
		{
			ans += count(lcm, a, b) * (n/lcm) + count(n%lcm, a, b);
		}
		else
		{
			ans = count(n, a, b);
		}
		
		printf("%I64d\n", ans);
	}

}