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  • 【BZOJ3160】万径人踪灭

    题面

    http://darkbzoj.tk/problem/3160

    题解

    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    #define ri register int
    #define N 500050
    #define mod 1000000007
    using namespace std;
    
    const double pi=acos(-1.0);
    struct fushu{double x,y;} A[N<<3],B[N<<3];
    fushu operator + (fushu a,fushu b) {return (fushu){a.x+b.x,a.y+b.y};}
    fushu operator - (fushu a,fushu b) {return (fushu){a.x-b.x,a.y-b.y};}
    fushu operator * (fushu a,fushu b) {return (fushu){a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};}
    
    int n,cnt,limit,r[N<<3],c[N<<3];
    int hw[N<<1];
    int tw[N<<3];
    char s0[N];
    char s1[N<<1];
    
    void FFT(fushu *a,int opt){
      for (ri i=0;i<limit;i++) if (i<r[i]) swap(a[i],a[r[i]]);
      for (ri mid=1;mid<limit;mid<<=1) {
        fushu wn=(fushu){cos(pi/mid),sin(pi/mid)*opt};
        for (ri j=0;j<limit;j+=2*mid) {
          fushu w0=(fushu){1,0};
          for (ri k=0;k<mid;k++,w0=w0*wn) {
            fushu x=a[j+k],y=w0*a[j+k+mid];
            a[j+k]=x+y; a[j+k+mid]=x-y;
          }
        }
      }
    }
    
    void init() {
      cnt=0;limit=1;
      while (limit<=4*n+2) limit<<=1,cnt++;
      for (ri i=0;i<limit;i++) r[i]=((r[i>>1]>>1)|((i&1)<<(cnt-1)));
    }
    
    void work(int ch) {
      for (ri i=0;i<limit;i++) A[i]=B[i]=(fushu){0.0,0.0};
      for (ri i=1;i<=2*n+1;i++) if (s1[i]==ch) A[i]=(fushu){1.0,0.0}; else A[i]=(fushu){0.0,0.0};
      for (ri i=1;i<=2*n+1;i++) B[i]=A[i];
      FFT(A,1); FFT(B,1);
      for (ri i=0;i<limit;i++) A[i]=A[i]*B[i];
      FFT(A,-1);
      for (ri i=0;i<limit;i++) c[i]+=(int)(A[i].x/limit+0.5);
    }
    
    int main(){
      tw[0]=1;
      for (ri i=1;i<(N<<3);i++) tw[i]=(tw[i-1]+tw[i-1])%mod;
      scanf("%s",s0+1);
      n=strlen(s0+1);
      for (ri i=1;i<=2*n+1;i++) if (i%2==0) s1[i]=s0[i/2]; else s1[i]='#';
      init();
      work('a');work('b');
      long long ans=0;
      for (ri i=1;i<=2*n+1;i++) {
        if (c[2*i]%2==0) c[2*i]/=2; else (c[2*i]+=1)/=2;
        ans+=tw[c[2*i]]-1;
        ans=(ans%mod+mod)%mod;
      }
      int mid=0,maxr=0;
      for (ri i=1;i<=2*n+1;i++) {
        if (i<=maxr) hw[i]=min(hw[2*mid-i],maxr-i);
        while (i+hw[i]+1<=2*n+1 && i-hw[i]-1>=1 && s1[i+hw[i]+1]==s1[i-hw[i]-1]) hw[i]++;
        if (i+hw[i]>maxr) maxr=i+hw[i],mid=i;
      }
      for (ri i=1;i<=2*n+1;i++) {
        if (s1[i]=='#') ans-=hw[i]/2; else ans-=(hw[i]+1)/2;
        ans=(ans%mod+mod)%mod;
      }
      cout<<ans<<endl;
    }
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  • 原文地址:https://www.cnblogs.com/shxnb666/p/11279718.html
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