很明显,求方程x*2^m=L (mod n+1)的解(1<=x<=n)

 1 var n,m,l,x,y,p:int64;
 2 function quick(x:int64):int64;
 3   var j:int64;
 4   begin
 5     j:=2; quick:=1;
 6     while x>0 do
 7     begin
 8       if x mod 2=1 then quick:=quick*j mod p;
 9       x:=x div 2;
10       j:=j*j mod p;
11     end;
12   end;
13 
14 procedure exgcd(a,b:int64;var x,y:int64);
15   var xx,yy:int64;
16   begin
17     if b=0 then
18     begin
19       x:=1;
20       y:=0;
21     end
22     else begin
23       exgcd(b,a mod b,xx,yy);
24       x:=yy;
25       y:=xx-(a div b)*yy;
26     end;
27   end;
28 
29 begin
30   readln(n,m,l);
31   p:=n+1;
32   m:=quick(m);
33   exgcd(m,p,x,y);
34   x:=(x+p) mod p;
35   writeln(qword(x)*qword(l) mod p);
36 end.
View Code

 

posted on 2015-04-10 14:09  acphile  阅读(129)  评论(0编辑  收藏  举报