【Luogu1919】 A*B Problem升级版（FFT）

题面戳我

题解

把每个数都直接看做一个多项式，每一位就是一项
现在求用FFT求出卷积
然后考虑一下进位就可以啦

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>#include<complex>#include<set>#include<map>#include<vector>using namespace std;#define MAX 140000complex<double> a[MAX],b[MAX];char s[MAX];int M,l;const double Pi=acos(-1);int r[MAX],N;int num[MAX];inline void Init(){    scanf("%d",&N);scanf("%s",s);N--;    for(int i=0;i<=N;++i)a[i]=s[N-i]-48;    scanf("%s",s);    for(int i=0;i<=N;++i)b[i]=s[N-i]-48;}void FFT(complex<double> *P,int opt){    for(int i=0;i<N;++i)if(i<r[i])swap(P[i],P[r[i]]);    for(int i=1;i<N;i<<=1)    {        complex<double> W(cos(Pi/i),opt*sin(Pi/i));        for(int p=i<<1,j=0;j<N;j+=p)        {            complex<double> w(1,0);            for(int k=0;k<i;k++,w*=W)            {                complex<double> X=P[j+k],Y=w*P[j+k+i];                P[j+k]=X+Y;P[j+k+i]=X-Y;            }        }    }}int main(){    Init();    M=N+N;    for(N=1;N<=M;N<<=1)l++;    for(int i=0;i<N;++i)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));    FFT(a,1);FFT(b,1);    for(int i=0;i<N;++i)a[i]=a[i]*b[i];    FFT(a,-1);    for(int i=0;i<N;++i)num[i]=(int)(a[i].real()/N+0.5);    //N=N+2;    for(int i=0;i<N;++i)num[i+1]+=num[i]/10,num[i]%=10;    while(!num[N-1])N--;    for(int i=N-1;i>=0;--i)printf("%d",num[i]);    return 0;    }