BZOJ 4259 残缺的字符串 ——FFT
Posted SfailSth ——Salt Fish
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了BZOJ 4259 残缺的字符串 ——FFT相关的知识,希望对你有一定的参考价值。
【题目分析】
同bzoj4503。
只是精度比较卡,需要试一试才能行O(∩_∩)O
用过long double,也加过0.4。最后发现判断的时候改成0.4就可以了
【代码】
#include <cmath> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define maxn 1200005 #define F(i,j,k) for (int i=j;i<=k;++i) #define D(i,j,k) for (int i=j;i>=k;--i) const double pi=acos(-1.0); struct Complex{ double x,y; Complex operator + (Complex a){Complex b; return b.x=x+a.x,b.y=y+a.y,b;} Complex operator - (Complex a){Complex b; return b.x=x-a.x,b.y=y-a.y,b;} Complex operator * (Complex a){Complex b; return b.x=x*a.x-y*a.y,b.y=x*a.y+y*a.x,b;} }a[maxn],b[maxn],c[maxn]; int A[maxn],B[maxn],la,lb,n,m=1,len,rev[maxn],ans[maxn],cnt=0; char s1[maxn],s2[maxn]; void FFT(Complex * x,int n,int f) { F(i,0,n-1) if (rev[i]>i) swap(x[rev[i]],x[i]); for (int m=2;m<=n;m<<=1) { int mid=m>>1; Complex wn; wn.x=cos(2.0*pi/m*f); wn.y=sin(2.0*pi/m*f); for (int i=0;i<n;i+=m) { Complex w; w.x=1.0; w.y=0; F(j,0,mid-1) { Complex a=x[i+j],b=x[i+j+mid]*w; x[i+j]=a+b; x[i+j+mid]=a-b; w=w*wn; } } } } int main() { // freopen("in.txt","r",stdin); scanf("%d%d",&lb,&la); scanf("%s",s2); scanf("%s",s1); n=la+lb+1; while (m<=n) m<<=1,len++; n=m; F(i,0,n-1) { int t=i,ret=0; F(j,1,len) ret<<=1,ret|=t&1,t>>=1; rev[i]=ret; } F(i,0,la-1) {A[i]=s1[i]-‘a‘+1; if (s1[i]==‘*‘) A[i]=0;} F(i,0,lb-1) {B[i]=s2[lb-1-i]-‘a‘+1; if (s2[lb-1-i]==‘*‘) B[i]=0;} F(i,0,n-1) a[i].x=A[i]*A[i]*A[i],a[i].y=0; F(i,0,n-1) b[i].x=B[i],b[i].y=0; FFT(a,n,1); FFT(b,n,1); F(i,0,n-1) c[i]=a[i]*b[i]; F(i,0,n-1) a[i].x=A[i],a[i].y=0; F(i,0,n-1) b[i].x=B[i]*B[i]*B[i],b[i].y=0; FFT(a,n,1); FFT(b,n,1); F(i,0,n-1) c[i]=c[i]+a[i]*b[i]; F(i,0,n-1) a[i].x=2*A[i]*A[i],a[i].y=0; F(i,0,n-1) b[i].x=B[i]*B[i],b[i].y=0; FFT(a,n,1); FFT(b,n,1); F(i,0,n-1) c[i]=c[i]-a[i]*b[i]; FFT(c,n,-1); F(i,0,n-1) c[i].x=c[i].x/n; // F(i,0,n-1) printf("%.3f ",c[i].x); printf("\n"); F(i,lb-1,la-1) if (fabs(c[i].x)<0.4) ans[++cnt]=i-lb+2; printf("%d\n",cnt); F(i,1,cnt) printf("%d%c",ans[i],i==cnt?‘\n‘:‘ ‘); }
以上是关于BZOJ 4259 残缺的字符串 ——FFT的主要内容,如果未能解决你的问题,请参考以下文章