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:=2; quick:=1;
    while x>0 do
    begin
      if x mod 2=1 then quick:=quick*j mod p;
      x:=x div 2;
      j:=j*j mod p;
    end;
  end;

procedure exgcd(a,b:int64;var x,y:int64);
  var xx,yy:int64;
  begin
    if b=0 then
    begin
      x:=1;
      y:=0;
    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+1;
  m:=quick(m);
  exgcd(m,p,x,y);
  x:=(x+p) mod p;
  writeln(qword(x)*qword(l) mod p);
end.