SGU 106 The equation(拓展欧几里得)

通解形式

然后利用x，y去求出范围，就能得到解的个数

注意特判a和b都为0的情况

代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

typedef long long ll;
ll a, b, c;
double a1, a2, b1, b2;

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 -= x * (a / b);
    return d;
}

ll solve() {
    ll x, y;
    if (a == 0 && b == 0) {
        if (c == 0) return (a2 - a1 + 1) * (b2 - b1 + 1);
        return 0;
    }
    ll d = exgcd(a, b, x, y);
    if (d < 0) d = -d;
    if (c % d) return 0;
    double l = -1e50, r = 1e50;
    if (b / d < 0) {
        r = min(r, floor((a1 - c / d * 1.0 * x) / (b / d)));
        l = max(l, ceil((a2 - c / d * 1.0 * x) / (b / d)));
    } else {
        l = max(l, ceil((a1 - c / d * 1.0 * x) / (b / d)));
        r = min(r, floor((a2 - c / d * 1.0 * x) / (b / d)));
    }
    if (a / d < 0) {
        l = max(l, ceil((c / d * 1.0 * y - b1) / (a / d)));
        r = min(r, floor((c / d * 1.0 * y - b2) / (a / d)));
    } else {
        r = min(r, floor((c / d * 1.0 * y - b1) / (a / d)));
        l = max(l, ceil((c / d * 1.0 * y - b2) / (a / d)));
    }
    return max(0LL, (ll)(r - l) + 1);
}

int main() {
    while (~scanf("%lld%lld%lld", &a, &b, &c)) {
        c = -c;
        scanf("%lf%lf", &a1, &a2);
        scanf("%lf%lf", &b1, &b2);
        printf("%lld\n", solve());
    }
    return 0;
}