FFT(快速傅立叶变换)：HDU 1402 A * B Problem Plus

Calculate A * B.

Input

Each line will contain two integers A and B. Process to end of file.

Note: the length of each integer will not exceed 50000.

Output

For each case, output A * B in one line.

Sample Input

1
2
1000
2

Sample Output

2
2000
　　
　　唉，模板题，膜的邝斌的模板。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const double PI = acos(-1.0);

struct complex{
    double r,i;
    complex(double r_=0.0,double i_=0.0)
    {
        r=r_;i=i_;
    }
    complex operator +(const complex &b)
    {
        return complex(r+b.r,i+b.i);
    }
    complex operator -(const complex &b)
    {
        return complex(r-b.r,i-b.i);
    }
    complex operator *(const complex &b)
    {
        return complex(r*b.r-i*b.i,i*b.r+r*b.i);
    }
};
 
void Rader(complex *a,int len)
{
    int k;
    for(int i=1,j=len/2;i<len-1;i++)
    {
        if(i<j)swap(a[i],a[j]);
        k=len/2;
        while(j>=k)
        {
            j-=k;
            k>>=1;
        }
        j+=k;
    }
}

void FFT(complex *a,int len,int on)
{
    Rader(a,len);
    for(int h=2;h<=len;h<<=1)
    {
        complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));
        for(int j=0;j<len;j+=h)
        {
            complex w(1,0);
            for(int k=j;k<j+h/2;k++)
            {
                complex u=a[k];
                complex v=a[k+h/2]*w;
                a[k]=u+v;
                a[k+h/2]=u-v;
                w=w*wn;
            }
        } 
    }
    if(on==-1)
        for(int i=0;i<len;i++)
            a[i].r/=len;
}

const int maxn = 200010;
complex Array1[maxn],Array2[maxn];
char str1[maxn],str2[maxn];
int sum[maxn],len,len1,len2;

int main()
{
    while(~scanf("%s%s",str1,str2))
    {
        len1=strlen(str1);
        len2=strlen(str2);    
        len=1;
        while(len<len1*2||len<len2*2)len<<=1;
        for(int i=0;i<len1;i++)
            Array1[i]=complex(str1[len1-i-1]-'0',0);
        for(int i=0;i<len2;i++)
            Array2[i]=complex(str2[len2-i-1]-'0',0);
        
        for(int i=len1;i<len;i++)
            Array1[i]=complex(0,0);
        for(int i=len2;i<len;i++)
            Array2[i]=complex(0,0);
        
        FFT(Array1,len,1);
        FFT(Array2,len,1);
        for(int i=0;i<len;i++)
            Array1[i]=Array1[i]*Array2[i];
        FFT(Array1,len,-1);
        memset(sum,0,sizeof(sum));    
        for(int i=0;i<len;i++){
            sum[i]+=(int)(Array1[i].r+0.5);
            sum[i+1]+=sum[i]/10;
            sum[i]%=10;
        }
        int p=len;
        while(!sum[p]&&p)p--;
        for(;p!=-1;p--)
            printf("%d",sum[p]);
        printf("
");                    
    }
    return 0;
}

 　　两个月后重打了一遍。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
const int maxn=400010;
const double PI=acos(-1.0);
struct complex{
    double r,i;
    complex(double r_=0.0,double i_=0.0){
        r=r_;i=i_;
    }
    complex operator +(complex a){
        return complex(r+a.r,i+a.i);
    }
    complex operator -(complex a){
        return complex(r-a.r,i-a.i);
    }
    complex operator *(complex a){
        return complex(r*a.r-i*a.i,r*a.i+a.r*i);
    }
}A[maxn],B[maxn];

void Rader(complex *a,int len){
    int k;
    for(int i=1,j=len>>1;i<len-1;i++){
        if(i<j)swap(a[i],a[j]);
        k=len>>1;
        while(j>=k){
            j-=k;
            k>>=1;
        }
        j+=k;
    }
}

void FFT(complex *a,int len,int on){
    Rader(a,len);
    for(int h=2;h<=len;h<<=1){
        complex wn(cos(-on*PI*2.0/h),sin(-on*PI*2.0/h));
        for(int j=0;j<len;j+=h){
            complex w(1,0);
            for(int k=j;k<j+(h>>1);k++){
                complex x=a[k];
                complex y=a[k+(h>>1)]*w;
                a[k]=x+y;
                a[k+(h>>1)]=x-y;
                w=w*wn;
            }    
        }
    }
    if(on==-1)
        for(int i=0;i<len;i++)
            a[i].r/=len;
    return;    
}

char s[maxn],t[maxn];
int ans[maxn];
int main(){
    while(~scanf("%s%s",s,t)){
    int lens=strlen(s);
    int lent=strlen(t),len=1;
    while(len<=lens+lent)len<<=1;
    memset(A,0,sizeof(A));
    memset(B,0,sizeof(B));
    for(int i=0;i<lens;i++)
        A[i].r=1.0*(s[lens-i-1]-'0');
    for(int i=0;i<lent;i++)
        B[i].r=1.0*(t[lent-i-1]-'0');
    FFT(A,len,1);FFT(B,len,1);
    for(int i=0;i<len;i++)
        A[i]=A[i]*B[i];
    FFT(A,len,-1);

    memset(ans,0,sizeof(ans));
    int in=0;
    for(int i=0;i<len;i++){
        ans[i]=in+floor(A[i].r+0.5);
        in=ans[i]/10;
        ans[i]%=10;
    }
    int i;
    for(i=len;i&&!ans[i];i--);
    while(~i)printf("%d",ans[i]),i--;
    printf("
");
    }
    return 0;
}