【bzoj3160】万径人踪灭 FFT

2023-11-10 21:40

文章标签 fft 万径人踪 bzoj3160

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【bzoj3160】万径人踪灭 FFT

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=3160

我是一个傻叉微笑脸

 1 #include<bits/stdc++.h>
 2 #define inf 1000000000
 3 #define ll long long
 4 #define N 200005
 5 #define mod 1000000007
 6 using namespace std;
 7 int read(){
 8   int x=0,f=1;char ch=getchar();
 9   while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
10   while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
11   return x*f;
12 }
13 struct CD{
14   double x,y;
15   CD(double a=0,double b=0){x=a;y=b;}
16   friend CD operator + (CD n1,CD n2){return CD(n1.x+n2.x,n1.y+n2.y);}
17   friend CD operator - (CD n1,CD n2){return CD(n1.x-n2.x,n1.y-n2.y);}
18   friend CD operator * (CD n1,CD n2){return CD(n1.x*n2.x-n1.y*n2.y,n1.x*n2.y+n1.y*n2.x);}
19 };
20 const double Pi=acos(-1.0);
21 int bit,n,nn,ans,m,mx,id,p[N<<1];
22 char s[N],st[N<<1];
23 CD a[N<<1],b[N<<1];
24 void FFT(CD *a,int n,int type){
25   for(int i=0,j=0;i<n;i++) {
26     if(j>i)swap(a[i],a[j]);
27     int k=n;
28     while(j&(k >>= 1))j&=~k;
29     j|=k;
30   }
31   for(int i=1;i<=bit;i++){
32     CD w_n(cos(2*type*Pi/(1<<i)),sin(2*type*Pi/(1<<i)));
33     for(int j=0;j<n;j+=(1<<i)){
34       CD w(1,0);
35       for(int k=j;k<j+(1<<(i-1));k++){
36         CD tmp=a[k],tt=w*a[k+(1<<(i-1))];
37         a[k]=tmp+tt;
38         a[k+(1<<(i-1))]=tmp-tt;
39         w=w*w_n;
40       }
41     }
42   }
43   if(type<0)for(int i=0;i<n;i++)a[i].x/=n;
44 }
45 int main(){
46   scanf("%s",s);
47   n=strlen(s);nn=n;
48   bit=1;
49   while((1<<bit)<(n<<1))bit++;
50   n=1<<bit;
51   for(int i=nn;i<n;i++)a[i]=b[i]=CD(0,0);
52   for(int i=0;i<nn;i++)if(s[i]=='a')a[i]=CD(1,0);
53   FFT(a,n,1);
54   for(int i=0;i<n;i++)a[i]=a[i]*a[i];
55   FFT(a,n,-1);
56  
57   for(int i=0;i<nn;i++)if(s[i]=='b')b[i]=CD(1,0);
58   FFT(b,n,1);
59   for(int i=0;i<n;i++)b[i]=b[i]*b[i];
60   FFT(b,n,-1);
61   
62   for(int i=0;i<n;i++){
63     int x=round(a[i].x+b[i].x);
64     x=(x+1)/2;
65     ans=(ans+(1<<x)-1)%mod;
66   }
67    
68   m=nn<<1|1;
69   for(int i=0;i<nn;i++)st[(i+1)<<1]=s[i];
70   mx=id=0;
71   for(int i=1;i<=m;i++){
72     if(mx>i)p[i]=min(p[2*id-i],mx-i);else p[i]=0;
73     while(i-p[i]-1>0&&i+p[i]+1<=m&&st[i-p[i]-1]==st[i+p[i]+1])p[i]++;
74     if(i+p[i]>mx)mx=i+p[i],id=i;
75     ans=((ans-(p[i]+1)/2)%mod+mod)%mod;
76   }
77   printf("%d\n",ans);
78   return 0;
79 }