【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=。=

　　

 1 /**************************************************************
 2     Problem: 2179
 3     User: Tunix
 4     Language: C++
 5     Result: Accepted
 6     Time:1236 ms
 7     Memory:9184 kb
 8 ****************************************************************/
 9  
10 //BZOJ 2179
11 #include<cmath>
12 #include<cstdio>
13 #include<cstring>
14 #include<cstdlib>
15 #include<iostream>
16 #include<algorithm>
17 #define rep(i,n) for(int i=0;i<n;++i)
18 #define F(i,j,n) for(int i=j;i<=n;++i)
19 #define D(i,j,n) for(int i=j;i>=n;--i)
20 using namespace std;
21 void read(int &v){
22     v=0; int sign=1; char ch=getchar();
23     while(ch<'0'||ch>'9'){ if (ch=='-') sign=-1; ch=getchar();}
24     while(ch>='0'&&ch<='9'){ v=v*10+ch-'0'; ch=getchar();}
25     v*=sign;
26 }
27 /******************tamplate*********************/
28 #define debug
29 const int N=150010;
30 const double pi=acos(-1.0);
31 struct comp{
32     double r,i;
33     comp(double _r=0.0,double _i=0.0):r(_r),i(_i){}
34 //  comp(){}
35     comp operator+(const comp &b)const{return comp(r+b.r,i+b.i);}
36     comp operator-(const comp &b)const{return comp(r-b.r,i-b.i);}
37     comp operator*(const comp &b)const{return comp(r*b.r-i*b.i,r*b.i+i*b.r);}
38 }a[N],b[N],c[N];
39  
40 void FFT(comp *a,int n,int type){
41     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
42         for(int s=n;j^=s>>=1,~j&s;);
43         if (i<j) swap(a[i],a[j]);
44     }
45     for(int m=1;m<n;m<<=1){
46         double u=pi/m*type; comp wm(cos(u),sin(u));
47         for(int i=0;i<n;i+=(m<<1)){
48             comp w(1,0);
49             rep(j,m){
50                 comp &A=a[i+j+m],&B=a[i+j],t=w*A;
51                 A=B-t; B=B+t; w=w*wm;
52             }
53         }
54     }
55     if (type==-1) rep(i,n) a[i].r/=n;
56 }
57 char s1[N],s2[N];
58 int ans[N];
59 int main(){
60     int n,k;
61     read(n);
62     scanf("%s%s",s1,s2);
63     rep(i,n){
64         a[i].r=s1[n-i-1]-'0';
65         b[i].r=s2[n-i-1]-'0';
66     }
67     for(k=1;k<=n*2;k<<=1);
68     FFT(a,k,1); FFT(b,k,1);
69     F(i,0,k) c[i]=a[i]*b[i];
70     FFT(c,k,-1);
71     F(i,0,n*2)
72         ans[i]=c[i].r+0.4;
73     int temp=0;
74     F(i,0,n*2){
75         if (ans[i]) temp=i;
76         ans[i+1]+=ans[i]/10;
77         ans[i]%=10;
78     }
79     D(i,temp,0) printf("%d",ans[i]);
80     return 0;
81 }
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posted @ 2015-01-21 21:58  Tunix  阅读(192)  评论(0)  编辑  收藏  举报