CodeForces - 1055C Lucky Days

最新推荐文章于 2020-12-25 16:00:27 发布

mmk27

最新推荐文章于 2020-12-25 16:00:27 发布

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分类专栏：思维数论

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本文探讨了两个竞争性程序员，Alice和Bob，在寻找连续幸运日子的最大数量过程中遇到的问题。通过分析各自的幸运和不幸日子周期，提出了一个算法解决方案，使用C++实现，以找出两人共同的最长连续幸运日。

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Bob and Alice are often participating in various programming competitions. Like many competitive programmers, Alice and Bob have good and bad days. They noticed, that their lucky and unlucky days are repeating with some period. For example, for Alice days [la;ra] are lucky, then there are some unlucky days: [ra+1;la+ta−1], and then there are lucky days again: [la+ta;ra+ta] and so on. In other words, the day is lucky for Alice if it lies in the segment [la+kta;ra+kta] for some non-negative integer k

The Bob's lucky day have similar structure, however the parameters of his sequence are different: lb

, rb, tb. So a day is a lucky for Bob if it lies in a segment [lb+ktb;rb+ktb], for some non-negative integer k

Alice and Bob want to participate in team competitions together and so they want to find out what is the largest possible number of consecutive days, which are lucky for both Alice and Bob.

Input

The first line contains three integers la

, ra, ta (0≤la≤ra≤ta−1,2≤ta≤109

) and describes Alice's lucky days.

The second line contains three integers lb

, rb, tb (0≤lb≤rb≤tb−1,2≤tb≤109

) and describes Bob's lucky days.

It is guaranteed that both Alice and Bob have some unlucky days.

Output

Print one integer: the maximum number of days in the row that are lucky for both Alice and Bob.

Examples

Input

0 2 5
1 3 5

Output

Input

0 1 3
2 3 6

Output

Note

The graphs below correspond to the two sample tests and show the lucky and unlucky days of Alice and Bob as well as the possible solutions for these tests.

题解：让 l 或 r 对其一个若无法对其那就两边取个最优解

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll cul(ll l1, ll r1, ll l2, ll r2, ll t) {
	l1 += t;
	r1 += t;
	return max(0ll, min(r1, r2) - max(l1, l2) + 1);
}

int main()
{
	ll l1, r1, v1, l2, r2, v2;
	while(~scanf("%lld %lld %lld %lld %lld %lld", &l1, &r1, &v1, &l2, &r2, &v2)) {
		l1++, r1++, l2++, r2++;
		ll d = __gcd(v1, v2);
		ll l = abs(r1 - r2);
		ll ans = 0;
		ll t1 = l / d * d;
		ll t2 = t1 + d;
		ans = max(ans, cul(l1, r1, l2, r2, t1));
		ans = max(ans, cul(l1, r1, l2, r2, t2));
		ans = max(ans, cul(l2, r2, l1, r1, t1));
		ans = max(ans, cul(l2, r2, l1, r1, t2));
		cout << ans << endl;
	}
	return 0;
}