【BZOJ】【2179】FFT快速傅里叶

FFT

　　做的第二道用到FFT的……好吧其实还是模板题-_-b

　　百度上说好像分治也能做……不过像FFT这种敲模板的还是省事=。=

　　

/**************************************************************
    Problem: 2179
    User: Tunix
    Language: C++
    Result: Accepted
    Time:1236 ms
    Memory:9184 kb
****************************************************************/
 
//BZOJ 2179
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define rep(i,n) for(int i=0;i<n;++i)
#define F(i,j,n) for(int i=j;i<=n;++i)
#define D(i,j,n) for(int i=j;i>=n;--i)
using namespace std;
void read(int &v){
    v=0; int sign=1; char ch=getchar();
    while(ch<'0'||ch>'9'){ if (ch=='-') sign=-1; ch=getchar();}
    while(ch>='0'&&ch<='9'){ v=v*10+ch-'0'; ch=getchar();}
    v*=sign;
}
/******************tamplate*********************/
#define debug
const int N=150010;
const double pi=acos(-1.0);
struct comp{
    double r,i;
    comp(double _r=0.0,double _i=0.0):r(_r),i(_i){}
//  comp(){}
    comp operator+(const comp &b)const{return comp(r+b.r,i+b.i);}
    comp operator-(const comp &b)const{return comp(r-b.r,i-b.i);}
    comp operator*(const comp &b)const{return comp(r*b.r-i*b.i,r*b.i+i*b.r);}
}a[N],b[N],c[N];
 
void FFT(comp *a,int n,int type){
    for(int i=1,j=0;i<n-1;++i){//只需改变1～n-2,0和n-1两个位置不变
        for(int s=n;j^=s>>=1,~j&s;);
        if (i<j) swap(a[i],a[j]);
    }
    for(int m=1;m<n;m<<=1){
        double u=pi/m*type; comp wm(cos(u),sin(u));
        for(int i=0;i<n;i+=(m<<1)){
            comp w(1,0);
            rep(j,m){
                comp &A=a[i+j+m],&B=a[i+j],t=w*A;
                A=B-t; B=B+t; w=w*wm;
            }
        }
    }
    if (type==-1) rep(i,n) a[i].r/=n;
}
char s1[N],s2[N];
int ans[N];
int main(){
    int n,k;
    read(n);
    scanf("%s%s",s1,s2);
    rep(i,n){
        a[i].r=s1[n-i-1]-'0';
        b[i].r=s2[n-i-1]-'0';
    }
    for(k=1;k<=n*2;k<<=1);
    FFT(a,k,1); FFT(b,k,1);
    F(i,0,k) c[i]=a[i]*b[i];
    FFT(c,k,-1);
    F(i,0,n*2)
        ans[i]=c[i].r+0.4;
    int temp=0;
    F(i,0,n*2){
        if (ans[i]) temp=i;
        ans[i+1]+=ans[i]/10;
        ans[i]%=10;
    }
    D(i,temp,0) printf("%d",ans[i]);
    return 0;
}