FFT
一直挺想学FFT。
1.n次单位根
http://wenku.baidu.com/link?url=x4Ot-WwMJOUCzs8soBnlVTQzlFJy0evAuVJYVG4Pe68ASMgws9LR1r7i02QZnx94JDSyoHNSb0s4dFObMyeTClbiNssF2dFqg5GZ0NVATJ_
2.FFT
http://blog.csdn.net/iamzky/article/details/22712347
然后。。。我发现。。。在网上找的资料我都看不懂。。。然后就看算法导论发现讲的非常详细。。。然后就看zky神犇的代码看看看看看看看看看看看看。。。。终于看懂了。。。据说还有NTT,CRT什么的。。。给我点智商行不行啊TAT。。。
以下是zky神犇的代码。
#include<cstdio> #include<cmath> #include<cstring> #include<iostream> #include<algorithm> using namespace std; const int maxn=1e6+10; struct cp{ double r,i; cp(double _r=0,double _i=0): r(_r),i(_i){} cp operator+(cp x){return cp(r+x.r,i+x.i);} cp operator-(cp x){return cp(r-x.r,i-x.i);} cp operator*(cp x){return cp(r*x.r-i*x.i,r*x.i+i*x.r);} }; cp a[maxn],b[maxn],A[maxn],x,y,c[maxn]; char s1[maxn],s2[maxn]; int sum[maxn],a1[maxn],a2[maxn],dig[maxn]; int len1,len2,rev[maxn],N,L; void FFT(cp a[],int flag){ for(int i=0;i<N;i++)A[i]=a[rev[i]]; for(int i=0;i<N;i++)a[i]=A[i]; for(int i=2;i<=N;i<<=1){ cp wn(cos(2*M_PI/i),flag*sin(2*M_PI/i)); for(int k=0;k<N;k+=i){ cp w(1,0); for(int j=k;j<k+i/2;j++){ x=a[j]; y=a[j+i/2]*w; a[j]=x+y; a[j+i/2]=x-y; w=w*wn; } } } if(flag==-1)for(int i=0;i<N;i++)a[i].r/=N; } int main(){ scanf("%s%s",s1,s2); len1=strlen(s1); len2=strlen(s2); for(N=1,L=0;N<max(len1,len2);N<<=1,L++);N<<=1;L++; for(int i=0;i<N;i++){ int len=0; for(int t=i;t;t>>=1)dig[len++]=t&1; for(int j=0;j<L;j++)rev[i]=(rev[i]<<1)|dig[j]; } for(int i=0;i<len1;i++)a1[len1-i-1]=s1[i]-'0'; for(int i=0;i<len2;i++)a2[len2-i-1]=s2[i]-'0'; for(int i=0;i<N;i++)a[i]=cp(a1[i]); for(int i=0;i<N;i++)b[i]=cp(a2[i]); FFT(a,1);FFT(b,1); for(int i=0;i<N;i++)c[i]=a[i]*b[i]; FFT(c,-1); for(int i=0;i<N;i++)sum[i]=c[i].r+0.5; for(int i=0;i<N;i++){ sum[i+1]+=sum[i]/10; sum[i]%=10; } int l=len1+len2-1; while(sum[l]==0&&l>0)l--; for(int i=l;i>=0;i--) putchar(sum[i]+'0'); putchar('\n'); return 0; }
ps:http://blog.miskcoo.com/2015/04/polynomial-multiplication-and-fast-fourier-transform 讲的也挺详细了。。。后面的扩展还没有看懂TAT
ps:NTT:快速数论变换:http://blog.csdn.net/acdreamers/article/details/39026505
ps:CRT:中国剩余定理: http://blog.csdn.net/acdreamers/article/details/8050018
51nod1028:修改后的FFT
#include<cstdio> #include<cstring> #include<cctype> #include<algorithm> #include<cmath> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) int read(){ int x=0;char c=getchar(); while(!isdigit(c)) c=getchar(); while(isdigit(c)) x=x*10+c-'0',c=getchar(); return x; } const int nmax=4e5+5; const int inf=0x7f7f7f7f; struct c{ double r,i; c(double r=0,double i=0):r(r),i(i){} c operator+(c x){ return c(r+x.r,i+x.i); } c operator-(c x){ return c(r-x.r,i-x.i); } c operator*(c x){ return c(r*x.r-i*x.i,r*x.i+i*x.r); } }; c a[nmax],b[nmax];int rev[nmax],ans[nmax],n;char sa[nmax],sb[nmax]; void fft(c *a,int f){ rep(i,0,n-1) if(i<rev[i]) swap(a[i],a[rev[i]]); for(int i=2;i<=n;i<<=1){ c wn(cos(2*M_PI/i),f*sin(2*M_PI/i)); for(int j=0;j<n;j+=i){ c w(1,0); for(int k=0;k<i/2;++k,w=w*wn){ c x=a[j+k],y=w*a[j+k+i/2]; a[j+k]=x+y;a[j+k+i/2]=x-y; } } } if(f<0) rep(i,0,n-1) a[i].r/=n; } int main(){ scanf("%s%s",sa,sb); int lena=strlen(sa)-1,lenb=strlen(sb)-1;n=max(lena,lenb); rep(i,0,lena) a[i]=c(sa[lena-i]-'0'); rep(i,0,lenb) b[i]=c(sb[lenb-i]-'0'); int m=n+n+2,L=0;for(n=1;n<=m;n<<=1,++L);//n是取不到的。那么1<<L就是取不到的。那么下一行应该是1<<L-1 rep(i,0,n) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1)); fft(a,1);fft(b,1); rep(i,0,n-1) a[i]=a[i]*b[i]; fft(a,-1); rep(i,0,m) ans[i]=a[i].r+0.1; rep(i,0,m) ans[i+1]+=ans[i]/10,ans[i]%=10; while(ans[m]==0&&m) --m; dwn(i,m,0) putchar(ans[i]+'0');putchar('\n'); return 0; }
bzoj2194:打一打模板
n=8 c[6]=a[7]b[1]+a[8]b[2]=a[7]b[7]+a[8]b[6] c[7]=a[8]b[1]...=a[8]b[7].所以将b逆向存一下就好了。速度进第一页啦嘿嘿嘿
#include<cstdio> #include<cstring> #include<cctype> #include<algorithm> #include<cmath> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) int read(){ int x=0;char c=getchar(); while(!isdigit(c)) c=getchar(); while(isdigit(c)) x=x*10+c-'0',c=getchar(); return x; } const int nmax=4e5+5; const int inf=0x7f7f7f7f; struct c{ double r,i; c(double r=0,double i=0):r(r),i(i){}; c operator+(c x){ return c(r+x.r,i+x.i); } c operator-(c x){ return c(r-x.r,i-x.i); } c operator*(c x){ return c(r*x.r-i*x.i,r*x.i+i*x.r); } }; c a[nmax],b[nmax]; int rev[nmax],ans[nmax],n; void fft(c *a,int f){ rep(i,0,n-1) if(i<rev[i]) swap(a[i],a[rev[i]]); for(int i=2;i<=n;i<<=1){ c wn(cos(2*M_PI/i),f*sin(2*M_PI/i)); for(int j=0;j<n;j+=i){ c w(1,0); for(int k=0;k<i/2;k++,w=w*wn){ c x=a[j+k],y=w*a[j+k+i/2]; a[j+k]=x+y;a[j+k+i/2]=x-y; } } } if(f<0) rep(i,0,n-1) a[i].r/=n; } int main(){ n=read();--n; rep(i,0,n) a[i]=c(read()),b[n-i]=c(read()); int m=n+n,L=0;for(n=1;n<=m;n<<=1,L++); rep(i,0,n) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1)); fft(a,1);fft(b,1); rep(i,0,n-1) a[i]=a[i]*b[i]; fft(a,-1); rep(i,m/2,m) printf("%d\n",(int)(a[i].r+0.1)); return 0; } /* 5 3 1 2 4 1 1 2 4 1 4 */
bzoj3527:力
话说还是不懂卷积是什么东西啊TAT。以下是我网上找到的神犇详细题解 by晴歌 http://www.cnblogs.com/Sunnie69/p/5574222.html
其实把m设成n+n+10一般就不会出现边界问题还有<=n不要是<n就可以了。话说这种写法优美多了
#include<cstdio> #include<cstring> #include<cctype> #include<algorithm> #include<cmath> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(int i=s;i>=t;i--) #define clr(x,c) memset(x,c,sizeof(x)) int read(){ int x=0;char c=getchar(); while(!isdigit(c)) c=getchar(); while(isdigit(c)) x=x*10+c-'0',c=getchar(); return x; } const int nmax=3e5+5;//三倍空间就可以了 const int inf=0x7f7f7f7f; struct c{ double r,i; c(double r=0,double i=0):r(r),i(i){} c operator+(c x){ return c(r+x.r,i+x.i); } c operator-(c x){ return c(r-x.r,i-x.i); } c operator*(c x){ return c(r*x.r-i*x.i,r*x.i+i*x.r); } }; c a[nmax],b[nmax];int rev[nmax],n; double f[nmax],g[nmax],t[nmax],ans[nmax]; void dft(c *a,int f){ rep(i,0,n) if(i<rev[i]) swap(a[i],a[rev[i]]); for(int i=2;i<=n;i<<=1){ c wn=c(cos(2*M_PI/i),f*sin(2*M_PI/i)); //因为这里忘了f*一直调TAT for(int j=0;j<n;j+=i){ c w=c(1,0); for(int k=0;k<i/2;++k,w=w*wn){ c x=a[j+k];c y=w*a[j+k+i/2]; a[j+k]=x+y;a[j+k+i/2]=x-y; } } } if(f<0) rep(i,0,n) a[i].r/=n; } void fft(double *ta,double *tb,c *a,c *b){ rep(i,0,n) a[i]=c(ta[i]),b[i]=c(tb[i]); dft(a,1);dft(b,1); rep(i,0,n) a[i]=a[i]*b[i]; dft(a,-1); } int main(){ n=read(); rep(i,1,n) scanf("%lf",&f[i]),g[i]=1.0/i/i; int m=n+n+10,L=0,tm=n;for(n=1;n<=m;n<<=1,L++); rep(i,0,n) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1)); fft(f,g,a,b); rep(i,0,tm) ans[i]=a[i].r,t[i]=f[tm-i]; fft(t,g,a,b); rep(i,0,tm) ans[i]-=a[tm-i].r; rep(i,1,tm) printf("%.3lf\n",ans[i]); return 0; } /* 5 4006373.885184 15375036.435759 1717456.469144 8514941.004912 1410681.345880 */