題目鏈接：洛谷、BZOJ2179

//將乘數拆成 a0*10^n + a1*10^(n-1) + ... + a_n-1的形式 
//可以發現多項式乘法就模擬了豎式乘法 所以用FFT即可 註意處理進位 
//n位*n位最多就只有2n位了 
//論putchar的速度。。還是快的 
#include <cmath>
#include <cstdio>
#include <cctype>
#include <algorithm>
#define gc() getchar()
const int N=6e4+5;
const double PI=acos(-1);

int
 
 n,ans[N<<1];
struct Complex
{
    double x,y;
    Complex(double xx=0,double yy=0) {x=xx, y=yy;}
    Complex operator + (const Complex &a) {return Complex(x+a.x, y+a.y);}
    Complex operator - (const Complex &a) {return Complex(x-a.x, y-a.y);}
    Complex operator * (const Complex &a) {return
 
 Complex(x*a.x-y*a.y, x*a.y+y*a.x);}
}A[N*3],B[N*3];

inline void read(Complex *a)
{
    int cnt=0;register char c=gc();
    for(;!isdigit(c);c=gc());
    for(;isdigit(c);a[cnt++].x=(double)(c-'0'),c=gc());
}
void Fast_Fourier_Transform(Complex *a,int lim,int opt)
{
    for(int j=0,i=0; i<lim; ++i)
    {
        if
 
(i>j) std::swap(a[i],a[j]);
        for(int l=lim>>1; (j^=l)<l; l>>=1);
    }
    for(int i=2; i<=lim; i<<=1)
    {
        int mid=i>>1;
        Complex Wn(cos(2.0*PI/i),opt*sin(2.0*PI/i)),t;
        for(int j=0; j<lim; j+=i)
        {
            Complex w(1,0);
            for(int k=0; k<mid; ++k,w=w*Wn)
                a[j+mid+k]=a[j+k]-(t=w*a[j+mid+k]),
                a[j+k]=a[j+k]+t;
        }
    }
}

int main()
{
    scanf("%d",&n);
    read(A), read(B);
    std::reverse(A,A+n), std::reverse(B,B+n);
    int lim=1;
    while(lim <= n<<1) lim<<=1;
    Fast_Fourier_Transform(A,lim,1);
    Fast_Fourier_Transform(B,lim,1);
    for(int i=0; i<=lim; ++i) A[i]=A[i]*B[i];
    Fast_Fourier_Transform(A,lim,-1);
    int t=n<<1;
    for(int i=0; i<=t; ++i) ans[i]=(int)(A[i].x/lim+0.5);
    for(int i=0; i<t; ++i) ans[i+1]+=ans[i]/10, ans[i]%=10;
    while(!ans[t] && t) --t;
    while(~t) putchar(ans[t--]+'0');

    return 0;
}

洛谷.1919.[模板]A乘B Problem升級版(FFT)