bzoj1965

很明显，求方程x*2^m=L (mod n+1)的解（1<=x<=n)

 var n,m,l,x,y,p:int64;
 function quick(x:int64):int64;
   var j:int64;
   begin
     j:=; quick:=;
     while x> do
     begin
       if x mod = then quick:=quick*j mod p;
       x:=x div ;
       j:=j*j mod p;
     end;
   end;

 procedure exgcd(a,b:int64;var x,y:int64);
   var xx,yy:int64;
   begin
     if b= then
     begin
       x:=;
       y:=;
     end
     else begin
       exgcd(b,a mod b,xx,yy);
       x:=yy;
       y:=xx-(a div b)*yy;
     end;
   end;

 begin
   readln(n,m,l);
   p:=n+;
   m:=quick(m);
   exgcd(m,p,x,y);
   x:=(x+p) mod p;
   writeln(qword(x)*qword(l) mod p);
 end.