FFT快速傅立叶

Description

给出两个n位10进制整数x和y，你需要计算x*y。

Input

第一行一个正整数n。第二行描述一个位数为n的正整数x。第三行描述一个位数为n的正整数y。

Output

输出一行，即x*y的结果。

Sample Input

1
3
4

Sample Output

12

数据范围：
n<=60000

HINT

 

Source

 

FFT裸题

具体见算导

code：

 1 #include<cstdio>

 2 #include<iostream>

 3 #include<cmath>

 4 #include<cstring>

 5 #include<algorithm>

 6 #define maxn 131072

 7 #define pi 3.14159265358979323846

 8 using namespace std;

 9 char ch;

10 int m,n,ans[maxn];

11 bool ok;

12 struct comp{

13     double rea,ima;

14     void clear(){rea=ima=0;}

15     comp operator +(const comp &x){return (comp){rea+x.rea,ima+x.ima};}

16     comp operator -(const comp &x){return (comp){rea-x.rea,ima-x.ima};}

17     comp operator *(const comp &x){return (comp){rea*x.rea-ima*x.ima,rea*x.ima+ima*x.rea};}

18 }a[maxn],b[maxn],c[maxn],tmp[maxn],w,wn;

19 void read(int &x){

20     for (ok=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=1;

21     for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());

22     if (ok) x=-x;

23 }

24 void read(comp *a){

25     for (int i=m-1;i>=0;i--){

26         for (ch=getchar();!isdigit(ch);ch=getchar());

27         a[i].rea=ch-'0';

28     }

29 }

30 void write(int *a){

31     int i;

32     for (i=n-1;i>=0&&!a[i];i--);

33     if (i<0){puts("0");return;}

34     for (;i>=0;i--) printf("%d",(int)a[i]);

35     puts("");

36 }

37 void fft(comp *a,int st,int siz,int step,int op){

38     if (siz==1) return;

39     fft(a,st,siz>>1,step<<1,op),fft(a,st+step,siz>>1,step<<1,op);

40     int x=st,x1=st,x2=st+step;

41     w=(comp){1,0},wn=(comp){cos(op*2*pi/siz),sin(op*2*pi/siz)};

42     for (int i=0;i<(siz>>1);i++,x+=step,x1+=(step<<1),x2+=(step<<1),w=w*wn)

43         tmp[x]=a[x1]+(w*a[x2]),tmp[x+(siz>>1)*step]=a[x1]-(w*a[x2]);

44     for (int i=st;siz;i+=step,siz--) a[i]=tmp[i];

45 }

46 int main(){

47     read(m),n=1;

48     while (n<(m<<1)) n<<=1;

49     read(a),read(b);

50     fft(a,0,n,1,1),fft(b,0,n,1,1);

51     for (int i=0;i<n;i++) c[i]=a[i]*b[i];

52     fft(c,0,n,1,-1);

53     for (int i=0;i<n;i++) ans[i]=(int)round(c[i].rea/(1.0*n));

54     for (int i=0;i<n;i++) ans[i+1]+=ans[i]/10,ans[i]%=10;

55     write(ans);

56     system("pause");

57     return 0;

58 }