第六周练习I题
I - 数论，线性方程
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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.
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
Sample Output
2 -3
sorry
7 -3
扩展欧几里德算法详解
http://www.cnblogs.com/hfc-xx/p/4744462.html
题解：紫书p313
扩展欧几里得算法是 用来在已知a,b 求解一组x,y使得x*a+y*b=gcd(a,b)
因为已知欧几里得算法gcd(a,b)=gcd(b,a%b) 所以x*a+y*b=gcd(a,b)=gcd(b,a%b)=x*b+y*a%b=x*b+y*(a-a/b*b)=y*a+(x-a/b*y)*b;
注意;a-a/b*b=a%b 这样就将a,b的线性组合化简b为a%b与的线性组合. 根据我的输出图可以看到： a,b都在减小，当b减小到0时， 我们就可以得出x=1,y=0; 然后递归回去就可以求出最终的x,y了
#include<iostream>
using namespace std;
void gcd(int a,int b,int & d,int &x,int &y)
{
if(!b)
{
d=a;x=;y=;
// cout<<d<<" "<<x<<" "<<y<<endl; //输出
}
else
{
gcd(b,a%b,d,y,x);
// cout<<b<<" "<<a%b<<" "<<d<<" "<<y<<" "<<x<<endl; //输出
y-=a/b*x;
// cout<<x<<" "<<y<<endl; //输出
}
}
int main()
{
int a,b,d,x,y;
while(cin>>a>>b)
{
gcd(a,b,d,x,y);
if(d!=) cout<<"sorry"<<endl;
else
{
while(x<) //x不能小于0
x+=b,y-=a;
cout<<x<<" "<<y<<endl;
}
}
return ;
}