HDU1402：A * B Problem Plus——题解

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
typedef double dl;
const dl pi=acos(-1.0);
const int N=2e5+10;
struct complex{//定义复数 
    dl x,y;
    complex(dl xx=0.0,dl yy=0.0){
        x=xx;y=yy;
    }
    complex operator +(const complex &b)const{
        return complex(x+b.x,y+b.y);
    }
    complex operator -(const complex &b)const{
        return complex(x-b.x,y-b.y);
    }
    complex operator *(const complex &b)const{
        return complex(x*b.x-y*b.y,x*b.y+y*b.x);
    }
};
void FFT(complex a[],int n,int on){
    for(int i=1,j=n>>1;i<n-1;i++){
        if(i<j)swap(a[i],a[j]);
        int k=n>>1;
        while(j>=k){j-=k;k>>=1;}
        if(j<k)j+=k;
    }
    for(int i=2;i<=n;i<<=1){
        complex res(cos(-on*2*pi/i),sin(-on*2*pi/i));
        for(int j=0;j<n;j+=i){
            complex w(1,0);
            for(int k=j;k<j+i/2;k++){
                complex u=a[k],t=w*a[k+i/2];
                a[k]=u+t;
                a[k+i/2]=u-t;
                w=w*res;
            }
        }
    }
    if(on==-1)
        for(int i=0;i<n;i++)a[i].x/=n;
}
char a[N],b[N];
complex x[N],y[N];
int ans[N];
int main(){
    while(cin>>a>>b){
        int len1=strlen(a),len2=strlen(b);
        int n=1;
        while(n<len1*2||n<len2*2)n<<=1;
        for(int i=0;i<len1;i++)x[i]=complex(a[len1-1-i]-‘0‘,0);
        for(int i=len1;i<n;i++)x[i]=complex(0,0);
        for(int i=0;i<len2;i++)y[i]=complex(b[len2-1-i]-‘0‘,0);
        for(int i=len2;i<n;i++)y[i]=complex(0,0);
        FFT(x,n,1);FFT(y,n,1);
        for(int i=0;i<n;i++)x[i]=x[i]*y[i];
        FFT(x,n,-1);
        for(int i=0;i<n;i++)ans[i]=(int)(x[i].x+0.5);
        for(int i=0;i<n;i++){
            ans[i+1]+=ans[i]/10;
            ans[i]%=10;
        }
        n=len1+len2-1;
        while(ans[n]<=0&&n>0)n--;
        for(int i=n;i>=0;i--)printf("%d",ans[i]);
        puts("");
    }
    return 0;
}