3123 高精度练习之超大整数乘法 - Wikioi

题目描述 Description

给出两个正整数A和B，计算A*B的值。保证A和B的位数不超过100000位。

输入描述 Input Description

读入两个用空格隔开的正整数

输出描述 Output Description

输出A*B的值

样例输入 Sample Input

4 9

样例输出 Sample Output

36

数据范围及提示 Data Size & Hint

两个正整数的位数不超过100000位

什么都不说了，裸FFT

看了一天的算导，勉强看懂了怎么O(nlogn)求dft，求dft^(-1)就实在看不懂了，唉，只好记下结论

首先是理解系数表达和点值表达，因为点值表达乘法是O（n）的，所以我们要通过系数表达算出点值表达，相乘后再算出点值表达

这个是分治求的dft奇偶分成两个序列，分别求出这两个序列的dft然后通过下面的代码就可以O（n）合并这个两个dft了

 for i:= to n>>(t+)- do
   begin
     p:=i<<(t+)+s;
     wt:=w[i<<t]*a[p+<<t];
     tt[i]:=a[p]+wt;
     tt[i+n>>(t+)]:=a[p]-wt;
   end;

先把单位根的几个性质搞懂，然后就基本上可以理解这个分治了（最好是自己手算一下......我就懒得说什么了，自己都不是很理解，代码还是抄别人的）

zky神犇手动搬运算导FFT

pascal代码（我从这里抄的）

 const
         s=;
 type
         cp=record
           x,y:double;
         end;
         arr=array[.. shl ]of cp;
 var
         c:array[..]of int64;
         a,b,w,tt:arr;
         n,i,p:longint;
         st:ansistring;
         wt:cp;

 operator *(var a,b:cp)c:cp;
 begin
         c.x:=a.x*b.x-a.y*b.y;
         c.y:=a.x*b.y+a.y*b.x;
 end;

 operator +(var a,b:cp)c:cp;
 begin
         c.x:=a.x+b.x;
         c.y:=a.y+b.y;
 end;

 operator -(var a,b:cp)c:cp;
 begin
         c.x:=a.x-b.x;
         c.y:=a.y-b.y;
 end;

 procedure dft(var a:arr;s,t:longint);
 begin
         if n>>t= then exit;
         dft(a,s,t+);
         dft(a,s+<<t,t+);
         for i:= to n>>(t+)- do
           begin
             p:=i<<(t+)+s;
             wt:=w[i<<t]*a[p+<<t];
             tt[i]:=a[p]+wt;
             tt[i+n>>(t+)]:=a[p]-wt;
           end;
         for i:= to n>>t- do
           a[i<<t+s]:=tt[i];
 end;

 procedure get(var a:arr);
 var
         i,l,ll:longint;
         c:char;
 begin
         read(c);
         st:='';
         while (ord(c)>=ord('')) and (ord(c)<=ord('')) do
           begin
             st:=st+c;
             read(c);
           end;
         while length(st)mod s<> do
           insert('',st,);
         ll:=length(st)div s;
         for l:= to ll do
           val(copy(st,l*s-s+,s),a[ll-l].x);
         while n>><l do
           n:=n<<;
 end;

 begin
         n:=;
         get(a);
         get(b);
         for i:= to n- do
           w[i].x:=cos(pi**i/n);
         for i:= to n- do
           w[i].y:=sin(pi**i/n);
         dft(a,,);
         dft(b,,);
         for i:= to n- do
           a[i]:=a[i]*b[i];
         for i:= to n- do
           w[i].y:=-w[i].y;
         dft(a,,);
         for i:= to n- do
           begin
             c[i]:=c[i]+round(a[i].x/n);
             c[i+]:=c[i] div ;
             c[i]:=c[i] mod ;
           end;
         while (c[i]=) and (i>) do
           dec(i);
         for p:=i downto  do
           begin
             str(c[p],st);
             while (i<>p) and (length(st)<s) do
               st:=''+st;
             write(st);
           end;
 end.

又写了一遍233

 const
     maxn=;
 type
     cp=record
         x,y:double;
     end;
     arr=array[..maxn]of cp;
 var
     a,b,w,tt:arr;
     n:longint;
     s:ansistring;
     c:array[..maxn]of longint;

 operator -(a,b:cp)c:cp;
 begin
     c.x:=a.x-b.x;
     c.y:=a.y-b.y;
 end;

 operator +(a,b:cp)c:cp;
 begin
     c.x:=a.x+b.x;
     c.y:=a.y+b.y;
 end;

 operator *(a,b:cp)c:cp;
 begin
     c.x:=a.x*b.x-a.y*b.y;
     c.y:=a.x*b.y+a.y*b.x;
 end;

 procedure get(var a:arr);
 var
     c:char;
     i,len:longint;
 begin
     s:='';
     read(c);
     while c in[''..''] do
         begin
             s:=s+c;
             read(c);
         end;
     len:=length(s);
     for i:= to len do
         a[len-i].x:=ord(s[i])-ord('');
     while n<len<< do
         n:=n<<;
 end;

 procedure dft(var a:arr;s,t:longint);
 var
     i,p:longint;
     wt:cp;
 begin
     if n>>t= then exit;
     dft(a,s+<<t,t+);dft(a,s,t+);
     for i:= to n>>t>>- do
         begin
             p:=i<<t<<+s;
             wt:=w[i<<t]*a[p+<<t];
             tt[i]:=a[p]+wt;
             tt[i+n>>t>>]:=a[p]-wt;
         end;
     for i:= to n>>t- do
         a[s+i<<t]:=tt[i];
 end;

 procedure main;
 var
     i:longint;
 begin
     n:=;get(a);get(b);
     for i:= to n- do w[i].x:=cos(pi**i/n);
     for i:= to n- do w[i].y:=sin(pi**i/n);
     dft(a,,);dft(b,,);
     for i:= to n- do a[i]:=a[i]*b[i];
     for i:= to n- do w[i].y:=-w[i].y;
     dft(a,,);
     for i:= to n- do c[i]:=round(a[i].x/n);
     for i:= to n- do
         begin
             inc(c[i+],c[i]div );
             c[i]:=c[i]mod ;
         end;
     i:=n-;
     while (c[i]=) and (i>) do dec(i);
     for i:=i downto  do write(c[i]);
 end;

 begin
     main;
 end.