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[Bzoj2242]常见数值算法

你被要求设计一个计算器完成以下三项任务:
1、给定y,z,p,计算Y^Z Mod P 的值;
2、给定y,z,p,计算满足xy≡ Z ( mod P )的最小非负整数;
3、给定y,z,p,计算满足Y^x ≡ Z ( mod P)的最小非负整数。
训练指南的题目,果然光看书不行,动手写才能掌握。
注意以下几点:
  • exgcd最后的gcd不需要取模
  • 负数的处理 (P+n) mod n
  • int64的使用(防止溢出)
(**************************************************************
    Problem: 2242
    User: HTwood
    Language: Pascal
    Result: Accepted
    Time:636 ms
    Memory:2568 kb
****************************************************************)
 
program math;
 
Const
  maxn=100000;
  maxm=2;
  StrErr='Orz, I cannot find x!';
 
Type
 rec=record
   val,ord:longint;
 end;
 
var
 p,q,n,m,op,gcd:int64;
 i:longint;
 t:array[0..maxn,0..maxm] of rec;
 
Function safe(P:int64):int64;inline;
  begin
  exit(((p+n) mod n+n) mod n);
end;
 
Function times(p,q:int64):int64;//P^Q (mod N)
  begin
  if q=0 then exit(1);
  times:=times(p,q shr 1);
  times:=safe(times*times);
  if odd(q) then times:=safe(times*p);
end;
 
Procedure Exgcd(a,b:int64;var x,y:int64);
  begin
  if b=0 then begin x:=1;y:=0;gcd:=a;exit; end;
  Exgcd(b,a mod b,y,x);
  y:=safe(y-a div b*x);
end;
 
Procedure Solve1;inline;//pX+Yn=Q
var
 x,y:int64;
  begin
  Exgcd(p,n,x,y);
  if q mod gcd<>0 then begin writeln(StrErr); exit; end;
  x:=safe(x);
  writeln(safe((q div gcd)*x));
end;
 
Procedure Addhash(val,ord:int64); inline;
var
 hash,i:longint;
  begin
  hash:=(val mod maxn+maxn) mod maxn;
  for i:=1 to maxm do
    begin
    if t[hash,i].ord>0 then continue;
    t[hash,i].ord:=ord;
    t[hash,i].val:=val;
    break;
  end;
end;
 
Function InHash(P:longint;var pos1,pos2:int64):boolean;inline;
var
 hash,i:longint;
  begin
  hash:=(p mod maxn+maxn) mod maxn;
  for i:=1 to maxm do
    begin
    if t[hash,i].ord=0 then break;
    if t[hash,i].val=p then
      begin
      pos1:=hash;
      pos2:=i;
      exit(true);
    end;
  end;
  exit(false);
end;
 
Function inv(P:int64):int64;inline;//pX=1 (mod n)
var
 x,y:int64;
  begin
  Exgcd(p,n,x,y);
  if gcd=1 then exit((x+n) mod n) else exit(-1);
end;
 
Procedure Solve2;inline;//P^x=q (mod n)
var
 size,v,now,pos1,pos2:int64;
 i:longint;
  begin
  q:=q mod n;
  size:=trunc(sqrt(n))+1;
  fillchar(t,sizeof(t),0);
  v:=inv(times(p,size));
  now:=1;
  for i:=1 to size do
    begin
    now:=(now*p) mod n;
    Addhash(now,i);
  end;
  For i:=0 to size-1 do
    begin
    if inhash(q,pos1,pos2) then
      begin
      writeln((size*i+t[pos1,pos2].ord) mod n);
      exit;
    end;
    q:=(q*v) mod n;
  end;
  Writeln(strErr);
end;
 
  begin
  readln(m,op);
  for i:=1 to m do
    begin
    readln(p,q,n);
    case op of
      1:writeln(times(p,q));
      2:solve1;
      3:solve2;
    end;
  end;
end.

 

posted on 2013-05-30 20:22  灰天飞雁  阅读(721)  评论(0编辑  收藏  举报

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