FFT模板(BZOJ2179)

实现了两个长度为n的大数相乘。

#include <cstdio>
#include <cmath>
#include <complex>
using namespace std;
#define pi acos(-1)

typedef complex<double> C;
const int N = ;
char s[N],t[N];
int n,m,l,r[N],c[N];
C a[N],b[N];

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

int main() {
    scanf("%d%s%s", &m, s, t);
    for(int i = ; i < m; i++) a[i] = s[m-i-]-'', b[i] = t[m-i-]-'';
    for(n = , m <<= ; n < m; n <<= ) l++;
    for(int i = ; i < n; i++) r[i] = (r[i>>]>>)|((i&)<<(l-));
    fft(a, ), fft(b, );
    for(int i = ; i < n; i++) a[i] *= b[i];
    fft(a, -);
    for(int i = ; i < n; i++) a[i] /= n;
    for(int i = ; i < m; i++) c[i] = (int)(a[i].real()+0.1);
    for(int i = ; i < m; i++) if(c[i] >= ) {
        c[i+] += c[i]/, c[i] %= ;
    } else if(!c[i] && i == m-) m--;
    for(int i = m-; ~i; i--) printf("%d", c[i]);
    return ;
}