fft模板 HDU 1402

// fft模板 HDU 1402

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <math.h>
#include <memory.h>
#include <bits/stdc++.h>
using namespace std;
#define LL long long
typedef pair<int,int> pii;
const LL inf = 0x3f3f3f3f;
const LL MOD =100000000LL;
const int N = 150010;
const double eps = 1e-8;
void fre() {freopen("in.txt","r",stdin);}
void freout() {freopen("out.txt","w",stdout);}
inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}return x*f;}


const double PI = acos(-1.0);
//复数结构体
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,r*b.i+i*b.r);
    }
};
/*
 * 进行FFT和IFFT前的反转变换。
 * 位置i和 （i二进制反转后位置）互换
 * len必须去2的幂
 */
void change(Complex y[],int len){
    int i,j,k;
    for(i = 1, j = len/2;i < len-1; i++){
        if(i < j)swap(y[i],y[j]);
        //交换互为小标反转的元素，i<j保证交换一次
        //i做正常的+1，j左反转类型的+1,始终保持i和j是反转的
        k = len/2;
        while( j >= k){
            j -= k;
            k /= 2;
        }
        if(j < k) j += k;
    }
}
/*
 * 做FFT
 * len必须为2^k形式，
 * on==1时是DFT，on==-1时是IDFT
 */
void fft(Complex y[],int len,int on){
    change(y,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 = y[k];
                Complex t = w*y[k+h/2];
                y[k] = u+t;
                y[k+h/2] = u-t;
                w = w*wn;
            }
        }
    }
    if(on == -1)
        for(int i = 0;i < len;i++)
            y[i].r /= len;
}

Complex a[N],b[N];
char s1[N/2],s2[N/2];
int ans[N];
int main(){
    while(~scanf("%s%s",s1,s2)){
         int len1=strlen(s1);
         int len2=strlen(s2);
         int l1=0,l2=0;
         while((1<<l1)<len1) l1++;
         while((1<<l2)<len2) l2++;
         int len=(1<<(max(l1,l2)+1));
         for(int i=0;i<len;i++){
            if(i<len1) a[i]=Complex(s1[len1-i-1]-'0',0);
            else a[i]=Complex(0,0);
            if(i<len2) b[i]=Complex(s2[len2-i-1]-'0',0);
            else b[i]=Complex(0,0);  
         }
         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);
         for(int i=0;i<len;i++)
            ans[i]=(int)(a[i].r+0.5);
         for(int i=0;i<len-1;i++){
            ans[i+1]+=ans[i]/10;
            ans[i]%=10;
         }
         bool flag=false;
         for(int i=len-1;i>=0;i--){
            if(ans[i]) printf("%d",ans[i]),flag=true;
            else if(flag||i==0)printf("0");
        }
        printf("
");
    }
    return 0;
}