NOIP 2012 同余方程

依据gcd(a,b)=gcd(b,a mod b);

设ax1+by1=gcd(a,b)=1

则bx2+(a mod b)y2=gcd(b,a mod b);

ax1+by1=bx2+(a mod b)y2;

ax1+by1=bx2+(a-[a/b]*b)y2=ay2+bx2-(a/b)*by2;

根据恒等定理得：x1=y2; y1=x2-[a/b]*y2;

 

 

var a,b,x,y:longint;
procedure gcd(a,b:longint);
      var t:longint;
begin
  if b=0 then
   begin
    x:=1;
    y:=0;
    exit;
   end else
   begin
    gcd(b,a mod b);
    t:=x;
    x:=y;
    y:=t-(a div b)*y;
   end;
end;

begin
 assign(input,'mod.in'); reset(input);
 assign(output,'mod.out'); rewrite(output);
 readln(a,b);
 gcd(a,b);
 while x<0 do x:=(x+b) mod b;
 write(x);
 close(input); close(Output);
end.