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

1 for i:=0 to n>>(t+1)-1 do

2   begin

3     p:=i<<(t+1)+s;

4     wt:=w[i<<t]*a[p+1<<t];

5     tt[i]:=a[p]+wt;

6     tt[i+n>>(t+1)]:=a[p]-wt;

7   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）

  1 const

  2         s=4;

  3 type

  4         cp=record

  5           x,y:double;

  6         end;

  7         arr=array[0..1 shl 17]of cp;

  8 var

  9         c:array[0..100000]of int64;

 10         a,b,w,tt:arr;

 11         n,i,p:longint;

 12         st:ansistring;

 13         wt:cp;

 14 

 15 operator *(var a,b:cp)c:cp;

 16 begin

 17         c.x:=a.x*b.x-a.y*b.y;

 18         c.y:=a.x*b.y+a.y*b.x;

 19 end;

 20 

 21 operator +(var a,b:cp)c:cp;

 22 begin

 23         c.x:=a.x+b.x;

 24         c.y:=a.y+b.y;

 25 end;

 26 

 27 operator -(var a,b:cp)c:cp;

 28 begin

 29         c.x:=a.x-b.x;

 30         c.y:=a.y-b.y;

 31 end;

 32 

 33 procedure dft(var a:arr;s,t:longint);

 34 begin

 35         if n>>t=1 then exit;

 36         dft(a,s,t+1);

 37         dft(a,s+1<<t,t+1);

 38         for i:=0 to n>>(t+1)-1 do

 39           begin

 40             p:=i<<(t+1)+s;

 41             wt:=w[i<<t]*a[p+1<<t];

 42             tt[i]:=a[p]+wt;

 43             tt[i+n>>(t+1)]:=a[p]-wt;

 44           end;

 45         for i:=0 to n>>t-1 do

 46           a[i<<t+s]:=tt[i];

 47 end;

 48 

 49 procedure get(var a:arr);

 50 var

 51         i,l,ll:longint;

 52         c:char;

 53 begin

 54         read(c);

 55         st:='';

 56         while (ord(c)>=ord('0')) and (ord(c)<=ord('9')) do

 57           begin

 58             st:=st+c;

 59             read(c);

 60           end;

 61         while length(st)mod s<>0 do

 62           insert('0',st,0);

 63         ll:=length(st)div s;

 64         for l:=1 to ll do

 65           val(copy(st,l*s-s+1,s),a[ll-l].x);

 66         while n>>1<l do

 67           n:=n<<1;

 68 end;

 69 

 70 begin

 71         n:=1;

 72         get(a);

 73         get(b);

 74         for i:=0 to n-1 do

 75           w[i].x:=cos(pi*2*i/n);

 76         for i:=0 to n-1 do

 77           w[i].y:=sin(pi*2*i/n);

 78         dft(a,0,0);

 79         dft(b,0,0);

 80         for i:=0 to n-1 do

 81           a[i]:=a[i]*b[i];

 82         for i:=0 to n-1 do

 83           w[i].y:=-w[i].y;

 84         dft(a,0,0);

 85         for i:=0 to n-1 do

 86           begin

 87             c[i]:=c[i]+round(a[i].x/n);

 88             c[i+1]:=c[i] div 10000;

 89             c[i]:=c[i] mod 10000;

 90           end;

 91         while (c[i]=0) and (i>0) do

 92           dec(i);

 93         for p:=i downto 0 do

 94           begin

 95             str(c[p],st);

 96             while (i<>p) and (length(st)<s) do

 97               st:='0'+st;

 98             write(st);

 99           end;

100 end.

View Code

又写了一遍233

 1 const

 2     maxn=400400;

 3 type

 4     cp=record

 5         x,y:double;

 6     end;

 7     arr=array[0..maxn]of cp;

 8 var

 9     a,b,w,tt:arr;

10     n:longint;

11     s:ansistring;

12     c:array[0..maxn]of longint;

13 

14 operator -(a,b:cp)c:cp;

15 begin

16     c.x:=a.x-b.x;

17     c.y:=a.y-b.y;

18 end;

19 

20 operator +(a,b:cp)c:cp;

21 begin

22     c.x:=a.x+b.x;

23     c.y:=a.y+b.y;

24 end;

25 

26 operator *(a,b:cp)c:cp;

27 begin

28     c.x:=a.x*b.x-a.y*b.y;

29     c.y:=a.x*b.y+a.y*b.x;

30 end;

31 

32 procedure get(var a:arr);

33 var

34     c:char;

35     i,len:longint;

36 begin

37     s:='';

38     read(c);

39     while c in['0'..'9'] do

40         begin

41             s:=s+c;

42             read(c);

43         end;

44     len:=length(s);

45     for i:=1 to len do

46         a[len-i].x:=ord(s[i])-ord('0');

47     while n<len<<1 do

48         n:=n<<1;

49 end;

50 

51 procedure dft(var a:arr;s,t:longint);

52 var

53     i,p:longint;

54     wt:cp;

55 begin

56     if n>>t=1 then exit;

57     dft(a,s+1<<t,t+1);dft(a,s,t+1);

58     for i:=0 to n>>t>>1-1 do

59         begin

60             p:=i<<t<<1+s;

61             wt:=w[i<<t]*a[p+1<<t];

62             tt[i]:=a[p]+wt;

63             tt[i+n>>t>>1]:=a[p]-wt;

64         end;

65     for i:=0 to n>>t-1 do

66         a[s+i<<t]:=tt[i];

67 end;

68 

69 procedure main;

70 var

71     i:longint;

72 begin

73     n:=1;get(a);get(b);

74     for i:=0 to n-1 do w[i].x:=cos(pi*2*i/n);

75     for i:=0 to n-1 do w[i].y:=sin(pi*2*i/n);

76     dft(a,0,0);dft(b,0,0);

77     for i:=0 to n-1 do a[i]:=a[i]*b[i];

78     for i:=0 to n-1 do w[i].y:=-w[i].y;

79     dft(a,0,0);

80     for i:=0 to n-1 do c[i]:=round(a[i].x/n);

81     for i:=0 to n-2 do

82         begin

83             inc(c[i+1],c[i]div 10);

84             c[i]:=c[i]mod 10;

85         end;

86     i:=n-1;

87     while (c[i]=0) and (i>0) do dec(i);

88     for i:=i downto 0 do write(c[i]);

89 end;

90 

91 begin

92     main;

93 end.

View Code