[FFT]luogu P1919 A*B Problem升级版

https://www.luogu.org/problemnew/show/P1919

分析

可以把一个数化为多项式的系数，如233

3+3x+2x^2

然后便可以进行快速傅里叶转换了

注意进位

 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <complex>
using namespace std;
typedef complex<double> cn;
const int N=262144;
const double pi=3.14159265358979;
int rev[N];
cn a[N],b[N];
int n,mx,bit;
bool p;

void Get_Rev(int mx) {
    for (int i=0;i<mx;i++)
        rev[i]=(rev[i>>1]>>1)|((i&1)<<bit-1);
}

void FFT (cn *a,int ir) {
    for (int i=0;i<mx;i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
    for (int i=1;i<mx;i<<=1) {
        cn wn=exp(cn(0,ir*pi/i));
        for (int len=i<<1,l=0;l<mx;l+=len) {
            cn wk=cn(1,0);
            for (int j=l;j<l+i;j++) {
                cn x=a[j],y=wk*a[j+i];
                a[j]=x+y;a[j+i]=x-y;
                wk*=wn;
            }    
        }
    }
    if (ir<0) for (int i=0;i<mx;i++) a[i]/=mx;
}

int main() {
    scanf("%d",&n);
    for (int i=0;i<n;i++) {
        char c;
        do {
            scanf("%c",&c);
        }
        while (c<'0'||c>'9');
        a[n-1-i]=(double)c-48;
    }
    for (int i=0;i<n;i++) {
        char c;
        do {
            scanf("%c",&c);
        }
        while (c<'0'||c>'9');
        b[n-1-i]=(double)c-48;
    }
    n--;
    mx=1;while (mx<=2*n) mx<<=1,bit++;
    Get_Rev(mx);
    FFT(a,1);FFT(b,1);
    for (int i=0;i<mx;i++) a[i]=a[i]*b[i];
    FFT(a,-1);
    mx=2*n;
    for (int i=0;i<=mx;i++) {
        a[i+1].real()+=(int)(a[i].real()+0.5)/10,a[i].real()=(int)(a[i].real()+0.5)%10;
        if (a[mx+1].real()>0) mx++;
    }
    for (int i=mx;i>=0;i--)
        if (p||a[i].real()>0) printf("%d",(int)(a[i].real()+0.5)),p=1;
}

View Code

 

转载于:https://www.cnblogs.com/mastervan/p/11112932.html