[FFT 模板题] HDU 1402:A*B Problem Plus

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题意

实现高精乘法。位数<=50000
题解

FFT 模板题。
#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const double PI=acos(-1.0);const int maxn=(1<<18)+5;struct E{    double real,imag;      E(double real=0,double imag=0):real(real),imag(imag){}    void operator /=(double x){ real/=x; imag/=x; }};E operator + (E &a,E &b){ return E(a.real+b.real,a.imag+b.imag); }  E operator - (E &a,E &b){ return E(a.real-b.real,a.imag-b.imag); }  E operator * (E &a,E &b){ return E(a.real*b.real-a.imag*b.imag,a.imag*b.real+a.real*b.imag); }int rev[maxn];void build_rev(int n){    rev[0]=0; int log2n=log2(n);    for(int i=1;i<=n-1;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(log2n-1));}const double eps=1e-10;int dcmp(double x){    if(fabs(x)<eps) return 0;    return x>0?1:-1;}int b[maxn];struct Complex_Line{    int n; E a[maxn];    void mem(){ for(int i=0;i<=n-1;i++) a[i].real=a[i].imag=0; n=0; }    void FFT(int k){        build_rev(n); for (int i=0;i<=n-1;i++) if (i<rev[i]) swap(a[i],a[rev[i]]);        for(int m=2;m<=n;m<<=1){            E wm(cos(2*PI/m),k*sin(2*PI/m));            for(int k=0;k<n-1;k+=m){                E w(1,0),t0,t1;                for(int j=0;j<=m/2-1;j++,w=w*wm) t0=a[k+j], t1=w*a[k+j+m/2], a[k+j]=t0+t1, a[k+j+m/2]=t0-t1;            }        }        if(k==-1) for(int i=0;i<=n;i++) a[i]/=n;    }    void write(){        memset(b,0,sizeof b);        for(int i=0;i<=n-1;i++) b[i]+=(int)(a[i].real+0.5), b[i+1]+=b[i]/10, b[i]%=10;        int m=n; while(m>1&&!b[m-1]) m--;        for(int i=m-1;i>=0;i--) printf("%d",b[i]);    }} A,B,C;char s[maxn];int main(){    freopen("hdoj1402.in","r",stdin);    freopen("hdoj1402.out","w",stdout);    for(scanf("%s",s);s[0]!='\000';s[0]='\000',scanf("%s",s)){        A.mem(); A.n=strlen(s); for(int i=0;i<=A.n-1;i++) A.a[i].real=s[A.n-1-i]-'0';        scanf("%s",s);         B.mem(); B.n=strlen(s); for(int i=0;i<=B.n-1;i++) B.a[i].real=s[B.n-1-i]-'0';        C.n=1; while(C.n<A.n+B.n) C.n<<=1;        A.n=C.n; A.FFT(1); B.n=C.n; B.FFT(1);        for(int i=0;i<=C.n-1;i++) C.a[i]=A.a[i]*B.a[i];        C.FFT(-1); C.write(); printf("\n");         }    return 0;}
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