SGU-106 The equation (Expanded Europe + detailed processing)

Portal

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Topic

给出$ax+by+c=0\; is$ a binary indefinite equation of the first degree, which satisfies{(x, y) ∣ x ∈ [lx, rx], y ∈ [ly, ry]} \{(x,y) | x \in [lx,rx] ,y\in [ly,ry]\}{ (x,y ) ∣ x∈[lx,rx],Y∈[ly,r y ] } The number of solutions.

Problem solving ideas

I thought I had written a lot of questions about expanding Europe and I knew it well, but this question hit my face again, $wa$ to doubt life...

First of all, it is not difficult to think that if $c$ moved to the right side of the equation is still negative, soccshould be$c$ changes sign, at the same time$a,b$ should also change sign, but at this time$a,$The sign of$b$ may be negative. At the beginning, my idea was to refer to the independent variables$x,y$ , that is,$|a|(-x)+|b|(-y)=c$ , and then take the negative directly after finding the special solution. Isn't it possible to do this? Yes, but not in this question. The reasons are as follows:

Because what we require is the number of solutions in a given solution space, suppose we don’t consider $a,$the symbol of$b$ , then the above equation is converted to$x=\frac{c-by}{a}$， This $a$ post sign change directly affects the final form of the general solution is changed, so if we want to$a$ negation, the correct way is given by the inverse of the solution space, i.e. becomes$[-rx,-Lx]$ for$y\; is\; the$ same.

Then after we solve the general solution, we need to get the number of solutions, because $x,$The transformation of$y$ is the opposite. It seems difficult to start, but in fact we only need to solve this inequality$lx\le x_0+k_1\frac{b}{gcd(a,b)}\le rx$ find$x$ in the solution space$k_1$The range of values, and then $y$ 's$k_2$For the value range of, just take one intersection of the two. At this time, you need to pay attention again. The left side should be rounded up, and the right side should be rounded down. Draw a picture to understand.

There is a special judgment $a=0,b=$Three cases of$0$

//
// Created by Happig on 2020/10/30
//
#include <bits/stdc++.h>
#include <unordered_map>
#include <unordered_set>

using namespace std;
#define fi first
#define se second
#define pb push_back
#define ins insert
#define Vector Point
#define ENDL "\n"
#define lowbit(x) (x&(-x))
#define mkp(x, y) make_pair(x,y)
#define mem(a, x) memset(a,x,sizeof a);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
const double eps = 1e-8;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double dinf = 1e300;
const ll INF = 1e18;
const int Mod = 1e9 + 7;
const int maxn = 2e5 + 10;

ll exgcd(ll a, ll b, ll &x, ll &y) {
    
    
    if (!b) {
    
    
        x = 1, y = 0;
        return a;
    }
    ll d = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}

ll gcd(ll a, ll b) {
    
    
    return b == 0 ? a : gcd(b, a % b);
}

int main() {
    
    
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    //ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    ll a, b, c, lx, rx, ly, ry;
    cin >> a >> b >> c >> lx >> rx >> ly >> ry;
    c = -c;
    if (c < 0) c = -c, a = -a, b = -b;
    if (a < 0) a = -a, swap(lx, rx), lx = -lx, rx = -rx;
    if (b < 0) b = -b, swap(ly, ry), ly = -ly, ry = -ry;
    if (a == 0 && b == 0) {
    
    
        if (c == 0) cout << (rx - lx + 1) * (ry - ly + 1) << endl;
        else cout << 0 << endl;
        return 0;
    } else if (a == 0) {
    
    
        if (c % b == 0 && ly <= c / b && c / b <= ry) cout << rx - lx + 1 << endl;
        else cout << 0 << endl;
        return 0;
    } else if (b == 0) {
    
    
        if (c % a == 0 && lx <= c / a && c / a <= rx) cout << ry - ly + 1 << endl;
        else cout << 0 << endl;
        return 0;
    }
    ll d = gcd(a, b);
    if (c % d) {
    
    
        cout << "0" << endl;
    } else {
    
    
        ll x, y;
        exgcd(a, b, x, y);
        x = x * c / d, y = y * c / d;
        //cout << x << " " << y << endl;
        a /= d, b /= d;
        int l = max(ceil(1.0 * (lx - x) / b), ceil(1.0 * (y - ry) / a)), r = min(floor(1.0 * (rx - x) / b),
                                                                                 floor(1.0 * (y - ly) / a));
        if (r >= l) cout << r - l + 1 << endl;
        else cout << 0 << endl;
    }
    return 0;
}