POJ 2891 Strange Way to Express Integers

求N个同余方程，用扩展欧几里得。

扩展欧几里得算法是用来在已知a,b的情况下求解一组x,y 使得 ax+by = gcd(a,b)

原理：

gcd(a,b) = gcd(b, a%b)

=> ax+by = gcd(a,b) = gcd(b, a%b);

=>ax+by=bx+(a%b)y;

=>ax+by=bx1+(a-(a/b)*b)y1=ay1+bx1-(a/b)*by1;

=>x=y1;y=x1-(a/b)*y1;

#include"iostream"
using namespace std;
__int64 egcd(__int64 a,__int64  b,__int64 &x,__int64 &y)//扩展欧几里得
{
	__int64 tmp,i;
	if(b==0)
	{
		x=1;y=0;
		return a;//返回gcd(m1,m2)
	}
	i=egcd(b,a%b,x,y);
	tmp=x;x=y;y=tmp-a/b*y;
	return i;//返回gcd(m1,m2)
}
int main()
{
	int n;
	while(cin>>n)
	{
		__int64 m1,r1,m2,r2,a,b,x,y;
		int flag=1,i,j;
		cin>>m1>>r1;
		for(i=1;i<n;i++)
		{
			cin>>m2>>r2;
			if(!flag)
				continue;
			a=egcd(m1,m2,x,y);
			if((r2-r1)%a!=0)
			{
				flag=0;
				continue;
			}
			x=x*(r2-r1)/a;
			b=m2/a;
			x=(x%b+b)%b;
			r1=r1+m1*x;
			m1=(m1*m2)/a;
		}
		if(flag)
			cout<<r1<<endl;
		else cout<<-1<<endl;
	}
	return 0;
}