扩展欧几里得的一些思考

假设方程为ax+by=c； 其中a和b的最大公倍数为d

1.在利用exgcd计算a和b的最大公约数时，如果a==0&&b==0，d=0，如果a和b其中一个等于0，比如说：a=0,b=4，则d=4

2.exgcd正确计算的条件是a和b都不为负所以遇到a或b是负数的情况时，对负数取相反数，然后将求得的结果取相反数

3.要注意一些特殊情况：

   (1)a=0,b=0时，只有c=0时，才能得到x=0,y=0的解

   (2)a=0时，只有c%b==0时才能得到解

   (3)b=0时，只有c%a==0时才能得到解

4.c%d!=0时，说明方程无解

一道例题：传送门：点击打开链接

There is an equation ax + by + c = 0. Given a,b,c,x1,x2,y1,y2 you must determine, how many integer roots of this equation are satisfy to the following conditions : x1<=x<=x2,   y1<=y<=y2. Integer root of this equation is a pair of integer numbers (x,y).

Input

Input contains integer numbers a,b,c,x1,x2,y1,y2 delimited by spaces and line breaks. All numbers are not greater than 10⁸ by absolute value.

Output

Write answer to the output.

Sample Input

1 1 -3
0 4
0 4

Sample Output

4

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<cmath>
using namespace std;
typedef long long ll;

ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    ll d = exgcd(b, a % b, x, y);
    ll tmp = x;
    x = y;
    y = tmp - a / b * y;
    return d;
}
int main()
{
    ll a, b, c, x1, x2, y1, y2;
    while(~scanf("%lld%lld%lld%lld%lld%lld%lld", &a, &b, &c, &x1, &x2, &y1, &y2))
    {
        c = -c;
        if(a == 0&& b == 0)
        {
            if(c != 0)
                printf("0\n");
            else printf("%lld\n", (x2-x1+1)*(y2-y1+1));
        }
        else if(a==0)
        {
            if(c%b==0)
            {
                if(y1>c/b||y2<c/b)
                    printf("0\n");
                else printf("%lld\n",x2-x1+1);
            }
        }
        else if(b==0)
        {
            if(c%a==0)
            {
                if(x1>c/a||x2<c/a)
                    printf("0\n");
                else printf("%lld\n",y2-y1+1);
            }
        }
        else
        {
            if(a < 0)
            {
                a = -a;
                ll tmp = x1;
                x1 = -x2;
                x2 = -tmp;
            }
            if(b < 0)
            {
                b = -b;
                ll tmp = y1;
                y1 = -y2;
                y2 = -tmp;
            }
            ll x, y;
            ll d = exgcd(a, b, x, y);
            if(c % d)
                printf("0\n");
            else
            {
                x = x*(c/d);
                y = y*(c/d);
                ll bd = b/d, ad = a/d;
                ll ri1 = (ll)floor((double)(x2-x)/(double)bd);
                ll ri2 = (ll)floor((double)(y-y1)/(double)ad);
                ll li1 = (ll)ceil((double)(x1-x)/(double)bd);
                ll li2 = (ll)ceil((double)(y-y2)/(double)ad);
                ll ri = min(ri1, ri2);
                ll li = max(li1, li2);
                if(ri < li)
                    printf("0\n");
                else printf("%lld\n", ri-li+1);
            }
        }

    }

}