LightOJ 1306 - Solutions to an Equation 裸EXGCD

本题是极其裸的EXGCD
AX+BY+C=0
给你a b c 和x与y的区间范围，问你整数解有几组
作为EXGCD入门，题目比较简单 主要需要考虑区间范围的向上、向下取整，及正负符号的问题
*问题是这正负号判断考虑让我WA无数次* **我好菜阿**

> 补充：关于使用扩展欧几里德算法解决不定方程的办法
>对于不定整数方程pa+qb=c，若 c mod Gcd(p, q)=0,则该方程存在整数解，否则不存在整数解。
>上面已经列出找一个整数解的方法，在找到p * a+q * b = Gcd(p, q)的一组解p0,q0后，p * a+q * b = Gcd(p, q)的其他整数解满足：
>p = p0 + b/Gcd(p, q) * t
>q = q0 – a/Gcd(p, q) * t(其中t为任意整数)
>至于pa+qb=c的整数解，只需将p * a+q * b = Gcd(p, q)的每个解乘上 c/Gcd(p, q) 即可
>——摘自他人博客

/** @Date    : 2016-10-21-15.24

  * @Author  : Lweleth (SoungEarlf@gmail.com)

  * @Link    : 

  * @Version : $

  */

#include <stdio.h>

#include <iostream>

#include <string.h>

#include <algorithm>

#include <utility>

#include <vector>

#include <map>

#include <set>

#include <math.h>

#include <string>

#include <stack>

#include <queue>

#define LL long long

#define MMF(x) memset((x),0,sizeof(x))

#define MMI(x) memset((x), INF, sizeof(x))

using namespace std;



const int INF = 0x3f3f3f3f;

const int N = 1e5+2000;





LL exgcd(LL a, LL b, LL &x, LL &y)

{

    LL d = a;

    if(b == 0)

    {

        x = 1;

        y = 0;

    }

    else

    {

        d = exgcd(b, a % b, y, x);

        y -= (a / b)*x;

    }

    return d;

}



int main()

{

    int T;

    int cnt = 0;

    cin >> T;

    while(T--)

    {

        LL a, b, c, x1, x2, y1, y2;

        scanf("%lld%lld%lld", &a, &b, &c);

        scanf("%lld%lld%lld%lld",&x1, &x2, &y1, &y2);

        c = -c;

        if(a < 0)

        {

            a = -a;

            swap(x1, x2);

            x1 = -x1;

            x2 = -x2;

        }

        if(b < 0)

        {

            b = -b;

            swap(y1, y2);

            y1 = -y1;

            y2 = -y2;

        }

        printf("Case %d: ", ++cnt);

        LL x0 = 0 , y0 = 0;





        LL g = exgcd(a, b, x0, y0);

        if(g!=0 && c % g != 0)

        {

            printf("0\n");

            continue;

        }

        if(a == 0 && b == 0)

        {

            if(!c)

                printf("%lld\n", (x2 - x1 + 1) * (y2 - y1 + 1));

            else printf("0\n");

            continue;



        }

        if(a == 0)

        {

            if(c / b >= y1 && c/b <= y2)

                printf("%lld\n", x2 - x1 + 1);

            else printf("0\n");

            continue;

        }

        if(b == 0)

        {

            if(c / a >= x1 && c / a <= x2)

                printf("%lld\n", y2 - y1 + 1);

            else printf("0\n");

            continue;



        }



        x0 = x0 * c / g;

        y0 = y0 * c / g;

        a /= g;

        b /= g;

        LL l = ceil((double)(x1 - x0)/(double)(b));

        LL r = floor((double)(x2 - x0)/(double)(b));

        LL w = ceil((double)(y0 - y2)/(double)(a));

        LL e = floor((double)(y0 - y1)/(double)(a));

        LL q = max(l, w);

        LL p = min(r, e);

        if(p < q)

            printf("0\n");

        else printf("%lld\n", p - q + 1);

    }

    return 0;

}