数论GCD——cf1055C

被一道数论题卡了半天

网上的题解说只要匹配l或者r就行，想了下还真是。。

能让r1和r2对其就让他们对其，不能对其就讨论一下两种情况就可以了

#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) + );
}

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 = ;
        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 ;
}