Week Five Practice I (extended Euclidean algorithm)

I-Number theory, linear equationsTime limit:1000MS Memory Limit:32768KB 64bit IO Format:%i64d & Amp %i64u

Description

The Sky is Sprite.  the Birds is Fly in the Sky.   the Wind is Wonderful.   blew Throw the Trees  Trees is Shaking, Leaves is falling.   Lovers Walk passing, and so is you.   ......... ............... Write in 中文版 class by Yifenfei 

Girls is clever and bright. In HDU every girl like math. Every girl like to solve math problem! Now, you have 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 , nonnegative integer A, b (0<a, b<=2^31)

Output

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

Sample Input

77 5110 4434 79

Sample Output

2-3sorry7-3Puzzle : Purple book p313The extended Euclidean algorithm is used to solve a set of X, Y, and X*A+Y*B=GCD (A, b) in a known A/b

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) because known Euclidean algorithm gcd (A, B) =gcd (B,A%B);

Note: A-a/b*b=a%b the linear combination of a A and a B as a linear combination of a%b. According to my output diagram can be seen: A, b are decreasing, when a reduced to 0 o'clock, we can draw x=1,y=0; And then you can go back and find the final x, Y.

#include <iostream>using namespacestd;voidgcdintAintBint& D,int&x,int&y) {    if(!b) {d=a;x=1; y=0; //cout<<d<< "" <<x<< "" <<y<<endl; //Output    }    Else{gcd (b,a%b,d,y,x); //cout<<b<< "<<a%b<<" "<<d<< " "<<y<<" <<x<<endl;//Outputy-=a/b*x; //cout<<x<< "" <<y<<endl; //Output    }}intMain () {intA,b,d,x,y;  while(cin>>a>>b) {gcd (a,b,d,x,y); if(d!=1) cout<<"Sorry"<<Endl; Else        {              while(x<0)//x cannot be less than 0x+=b,y-=A; cout<<x<<" "<<y<<Endl; }    }    return 0;}

Week Five Practice I (extended Euclidean algorithm)