BZOJ2179:FFT快速傅立叶
https://www.lydsy.com/JudgeOnline/problem.php?id=2179
FFT模板题
1 #include<cmath> 2 #include<cstdio> 3 #include<algorithm> 4 using namespace std; 5 const int maxn=140005; 6 const double pi=acos(-1); 7 int n,m=1,lg; 8 char s1[maxn],s2[maxn]; 9 int rev[maxn],ans[maxn]; 10 struct complex{ 11 double x,y; 12 complex(){}; 13 complex(double xx,double yy){x=xx,y=yy;} 14 complex operator+(const complex &a)const{return complex(a.x+x,a.y+y);} 15 complex operator-(const complex &a)const{return complex(x-a.x,y-a.y);} 16 complex operator*(const complex &a)const{return complex(x*a.x-y*a.y,x*a.y+y*a.x);} 17 }a[maxn],b[maxn],c[maxn]; 18 void FFT(complex *a,int len,int sign){ 19 for(int i=0;i<len;i++)if(rev[i]<i)swap(a[i],a[rev[i]]); 20 for(int i=2;i<=len;i<<=1){ 21 complex wn(cos(2*pi/i*sign),sin(2*pi/i*sign)); 22 for(int j=0;j<len;j+=i){ 23 complex w(1,0); 24 for(int k=0;k<(i>>1);k++,w=w*wn){ 25 complex x=a[j+k],y=w*a[j+(i>>1)+k]; 26 a[j+k]=x+y,a[j+k+(i>>1)]=x-y; 27 } 28 } 29 } 30 if(sign==-1)for(int i=0;i<len;i++)a[i].x/=len; 31 } 32 int main(){ 33 scanf("%d",&n); 34 scanf("%s%s",s1,s2); 35 for(int i=n-1;~i;i--)a[i].x=s1[n-1-i]-'0'; 36 for(int i=n-1;~i;i--)b[i].x=s2[n-1-i]-'0'; 37 n=n*2; 38 for(;m<=n;lg++,m<<=1); 39 for(int i=0;i<m;i++)rev[i]=(rev[i>>1]>>1)|((i&1)<<(lg-1)); 40 FFT(a,m,1),FFT(b,m,1); 41 for(int i=0;i<m;i++)c[i]=a[i]*b[i]; 42 FFT(c,m,-1); 43 for(int i=0;i<m;i++)ans[i]=(int)(c[i].x+0.5); 44 for(int i=0;i<m;i++)ans[i+1]+=ans[i]/10,ans[i]%=10; 45 if(ans[m])m++; 46 while(!ans[m]&&m)m--; 47 for(int i=m;~i;i--)printf("%d",ans[i]); 48 return 0; 49 }
