bzoj千题计划166：bzoj2179: FFT快速傅立叶

http://www.lydsy.com/JudgeOnline/problem.php?id=2179

FFT做高精乘

#include<cmath>
#include<cstdio>
#include<complex>

using namespace std;

#define N 60001
const int M=(<<)+;

const double pi=acos(-);

typedef complex<double> E;

int n;
char s[N];

int r[M];

E a[M],b[M];

int res[M];

void fft(E *a,int f)
{
    for(int i=;i<n;++i)
        if(i<r[i]) swap(a[i],a[r[i]]);
    for(int i=;i<n;i<<=)
    {
        E wn(cos(pi/i),f*sin(pi/i));
        for(int p=i<<,j=;j<n;j+=p)
        {
            E w(,);
            for(int k=;k<i;++k,w*=wn)
            {
                E x=a[j+k],y=w*a[j+k+i];
                a[j+k]=x+y; a[j+k+i]=x-y;
            }
        }
    }
}

int main()
{
    int m;
    scanf("%d",&n);
    n--; m=n;
    scanf("%s",s);
    for(int i=n;i>=;--i) a[n-i]=s[i]-'';
    scanf("%s",s);
    for(int i=m;i>=;--i) b[m-i]=s[i]-'';
    m+=n; int bit=;
    for(n=;n<=m;n<<=) bit++;
    for(int i=;i<n;++i) r[i]=(r[i>>]>>)|((i&)<<bit-);
    fft(a,);
    fft(b,);
    for(int i=;i<n;++i) a[i]=a[i]*b[i];
    fft(a,-);
    for(int i=;i<=m;++i) res[i]=(int)(a[i].real()/n+0.5);
    for(int i=;i<=m;++i) res[i+]+=res[i]/,res[i]%=;
    if(res[m+]) m++;
    //while(m && !res[m]) --m;
    for(int i=m;i>=;--i) printf("%d",res[i]);
}