扩展欧几里得算法

紫书：

求直线ax+by+c=0上的整数点（x，y）；

先求ax+by+c=gcd(a,b)的解

void gcd(int a,int b,int &d,int &x,int &y)//d为a，b的最大公因数
{
if(!b)
{
d=a;
x=1;
y=0;
}
else
{
gcd(b,a%b,d,y,x);
y-=x*(a/b);
}
}

若方程的一组解为（x0,y0)，则通解为（x0+kb',y0+ka')（a'=a/gcd(a,b),b'=b/gcd(a,b)）;k为任意整数

例题：

Now tell you two nonnegative integer a and b. Find the nonnegative integer X and integer Y to satisfy X*a + Y*b = 1. If no such answer print "sorry" instead.

 

Input

The input contains multiple test cases.

Each case two nonnegative integer a,b (0<a, b<=2^31)

 

Output

output nonnegative integer X and integer Y, if there are more answers than the X smaller one will be choosed. If no answer put "sorry" instead.

 

Sample Input

77 51
10 44
34 79

#include<cstdio>
#include<iostream>
using namespace std;
void gcd(int a,int b,int &d,int &x,int &y)
{
if(!b)
{
d=a;
x=1;
y=0;
}
else
{
gcd(b,a%b,d,y,x);
y-=x*(a/b);
}
}
int main()
{
int x,y,a,b,d;
<
4000
span style="font-size:18px;"> while(~scanf("%d%d",&a,&b))
{
gcd(a,b,d,x,y);
if(d!=1)
{
printf("sorry\n");
continue;
}
else
{
while(x<=0)
{
x=x+b;
y=y-a;
}
printf("%d %d\n",x,y);
}
}
return 0;
}