Description
The Sky is Sprite.
The Birds is Fly in the Sky.
The Wind is Wonderful.
Blew Throw the Trees
Trees are Shaking, Leaves are Falling.
Lovers Walk passing, and so are You.
................................Write in English class by yifenfei
Girls are clever and bright. In HDU every girl like math. Every girl like to solve math problem!
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.
The Birds is Fly in the Sky.
The Wind is Wonderful.
Blew Throw the Trees
Trees are Shaking, Leaves are Falling.
Lovers Walk passing, and so are You.
................................Write in English class by yifenfei
Girls are clever and bright. In HDU every girl like math. Every girl like to solve math problem!
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)
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
Sample Output
2 -3 sorry 7 -3
解题思路：这个问题我本来是用公式将y求出来然后看它是不是整数，虽然答案对了，但是提交总是超时。所以看了书上的扩展欧几里德算法之后才有了一点眉目。
程序代码：
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
long long gcd(long long a,long long b,long long &x,long long &y)
{
if(b==0) {
x=1;y=0;
return a;
}
else{
long long d=gcd(b,a%b,x,y);
long long temp=x;
x=y;
y=temp-a/b*y;
return d;
}
}
int main()
{
long long x,y,a,b;
while(cin>>a>>b)
{
long long ans=gcd(a,b,x,y);
if(1%ans)
printf("sorry
");
else{
while(x<0)
x+=b,y-=a;
printf("%d %d
",x,y);
}
}
return 0;
}