SGU 106 The Equation 扩展欧几里得应用

Sol：线性不定方程+不等式求解

证明的去搜下别人的证明就好了。。。数学题。

 

#include <algorithm>

#include <cstdio>

#include <iostream>



using namespace std;



long long extend_gcd(long long a,long long b,long long &x,long long &y)

{

	if(a==0&&b==0) return -1;

	if(b==0){x=1;y=0;return a;}

	long long d=extend_gcd(b,a%b,y,x);

	y-=a/b*x;

	return d;

}



long long a,b,c,x1,x2,y1,y2,ans;



int main()

{

	cin>>a>>b>>c>>x1>>x2>>y1>>y2;

	c=-c;

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

	{

		if(c==0)  ans=(x2-x1+1)*(y2-y1+1);

	}

	else if(a==0)

	{

		if(c%b==0&&c/b>=y1&&c/b<=y2)

			ans=x2-x1+1;

	}

	else if(b==0)

	{

		if(c%a==0&&c/a>=x1&&c/a<=x2)

			ans=y2-y1+1;

	}

	else

	{

		long long x=0,y=0;

		long long d=extend_gcd(a,b,x,y);

		if(c%d==0)

		{

			long long X=x*(c/d);

			long long Y=y*(c/d);

			long long k1,k2,k3,k4;

			if(x1<=X||(x1-X)*d%b==0)

				k1=(x1-X)*d/b;

			else

				k1=(x1-X)*d/b+1;

			if(x2>=X||(X-x2)*d%b==0)

				k2=(x2-X)*d/b;

			else

				k2=(x2-X)*d/b-1;

			if(y1<=Y||(y1-Y)*d%a==0)

				k3=(Y-y1)*d/a;

			else

				k3=(Y-y1)*d/a-1;

			if(y2>=Y||(Y-y2)*d%a==0)

				k4=(Y-y2)*d/a;

			else

				k4=(Y-y2)*d/a+1;

			if(k1>k2) swap(k1,k2);

			if(k3>k4) swap(k3,k4);

			ans=min(k2,k4)-max(k1,k3)+1;

		}

	}

	printf("%I64d\n",ans);

	return 0;

}