sgu106（擴展歐幾里得）

給你7個數，a,b,c,x1,x2,y1,y2.求滿足a*x+b*y=-c的解x滿足x1<=x<=x2,y滿足y1<=y<=y2.求滿足條件的解的個數

用擴展歐幾里得求得一組解，然後用特解表示通解，這樣解出根據x1，x2，y1，y2，解出k，解出k的時候注意變號，也要注意a=0或者b=0,或者c=0的情況，

#include<set>
#include<queue>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<utility>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define MM 1200010
#define Inf (1<<30)
#define LL long long
#define MOD 1000000009
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
LL ceil_div(LL x,LL y)
{
    return ceil(double(x)/double(y));
}
LL floor_div(LL x,LL y)
{
    return floor(double(x)/double(y));
}
LL exgcd(LL a, LL b, LL &x, LL &y)
{
	if (b == 0)
	{
		x = 1; y = 0;
		return a;
	}
	LL g = exgcd(b, a%b, x, y);
	LL u = x;
	x = y;
	y = u - a / b*y;
	return g;
}
int main()
{
	LL a, b, c;
	LL xx1, xx2, yy1, yy2;
	//freopen("D:\\oo.txt", "r", stdin);
	while (~scanf("%lld%lld%lld%lld%lld%lld%lld", &a, &b, &c, &xx1, &xx2, &yy1, &yy2))
	{
		if(a==0&&b==0){
            if(c==0)printf("%lld\n", (yy2 - yy1+1)*( xx2 - xx1+1));
            else puts("0");
		}
		else if (b == 0 && a != 0)
		{
			if (c%a==0&&xx1 <= -c / a&&-c / a <= xx2)
				printf("%lld\n", yy2 - yy1+1);
			else puts("0");
		}
		else if (a == 0 && b != 0)
		{
			if (c%b==0&&yy1 <= -c / b&&-c / b <= yy2)printf("%lld\n", xx2 - xx1+1);
			else puts("0");
		}
		else
		{
		    LL x, y;
            LL g = exgcd(a, b, x, y);
            if (-c%g != 0)puts("0");
            else {
                LL d=-c/g;
                a/=g;b/=g;
                x*=d;y*=d;
                LL X1,Y1,X2,Y2;
                if(b>0)X1=ceil_div(xx1-x,b),Y1=floor_div(xx2-x,b);
                else X1=ceil_div(xx2-x,b),Y1=floor_div(xx1-x,b);

                if(a>0)X2=ceil_div(y-yy2,a),Y2=floor_div(y-yy1,a);
                else X2=ceil_div(y-yy1,a),Y2=floor_div(y-yy2,a);

                LL res=min(Y1,Y2)-max(X1,X2)+1;
                if(res>=0)printf("%lld\n",res);
                else puts("0");
            }
		}
	}
	return 0;
}