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  • LightOJ

    Ax + By + C = 0,已知(x in[x1,x2], y in [y1,y2])
    求解的个数

    求出一组特解之后,$$x = frac{c}{d}x_0 + kfrac{b}{d}, y = frac{c}{d}y_0 - kfrac{a}{d} (k in Z)$$求出k的范围即可
    (x1 ≤ x + kfrac{b}{d} ≤x2)
    (d * b > 0:(x1 - x) frac{d}{b}≤k≤(x2 - x)frac{d}{b})
    (d*b < 0:(x2 - x) frac{d}{b} ≤k≤(x1 - x)frac{d}{b})
    同理对于y
    (y1 ≤y - kfrac{d}{a}≤y2)
    (d * a>0:(y - y2)frac{d}{a} ≤k≤(y - y1) frac{d}{a})
    (d * a<0:(y - y1)frac{d}{a} ≤k≤(y - y2) frac{d}{a})
    特判(a = 0,b = 0,c = 0)(a = 0,b != 0)(a != 0,b = 0)

    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #define ll long long
    using namespace std;
    void ex_gcd(ll a, ll b, ll &d, ll &x, ll &y){
        if(b == 0){
            d = a, x = 1, y = 0;
            return;
        }
        ex_gcd(b, a % b, d, y, x);
        y -= x * (a / b);
    }
    void solve(ll a, ll b, ll c, ll x1, ll x2, ll y1, ll y2){
        ll d, x, y;
        ex_gcd(a, b, d, x, y);
        if(a == 0 && b == 0){
            if(c == 0)printf("%lld
    ", (x2 - x1 + 1) * (y2 - y1 + 1));
            else printf("0
    ");
            return;
        }
        if(a == 0){
            ll tmp = c / b;
            if(tmp >= y1 && tmp <= y2 && c % b == 0)
                printf("%lld
    ", x2 - x1 + 1);
            else printf("0
    ");
            return;
        }
        if(b == 0){
            ll tmp = c / a;
            if(tmp >= x1 && tmp <= x2 && c % a == 0)
                printf("%lld
    ", y2 - y1 + 1);
            else printf("0
    ");
            return;
        }
        if(c % d != 0){
            printf("0
    ");
        }else{
            x = c / d * x, y = c / d * y;
            ll k, t_x1, t_x2, t_y1, t_y2;
            if(d * b > 0){
                t_x1 = ceil((x1 - x) * 1.0 * d / b);
                t_x2 = floor((x2 - x) * 1.0 * d / b);
            }else{
                t_x1 = ceil((x2 - x) * 1.0 * d / b);
                t_x2 = floor((x1 - x) * 1.0 * d / b);
            }
            if(d * a > 0){
                t_y1 = ceil((y - y2) * 1.0 * d / a);
                t_y2 = floor((y - y1) * 1.0 * d / a);
            }else{
                t_y1 = ceil((y - y1) * 1.0 * d / a);
                t_y2 = floor((y - y2) * 1.0 * d / a);
            }
            ll l = max(t_x1, t_y1), r = min(t_x2, t_y2);
            printf("%lld
    ", max(0ll, r - l + 1));
        }
    }
    int main(){
        int t;
        scanf("%d", &t);
        for(int i = 1; i <= t; i++){
            ll a, b, c, x1, x2, y1, y2;
            scanf("%lld%lld%lld", &a, &b, &c);
            scanf("%lld%lld%lld%lld", &x1, &x2, &y1, &y2);
            printf("Case %d: ", i);
            solve(a, b, -c, x1, x2, y1, y2);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Emcikem/p/12968656.html
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