# bzoj3160 万径人踪灭

FFT加上一些计数原理即可A掉此题,主要需要注意的是两字符间的位置也需要运算.

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<cmath>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<iomanip>
using namespace std;
#define LL long long
#define up(i,j,n) for(LL i=j;i<=n;i++)
#define pii pair<LL,LL>
#define db double
#define eps 1e-4
#define FILE "dealing"
LL x=0,f=1,ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch<='9'&&ch>='0'){x=(x<<1)+(x<<3)+ch-'0',ch=getchar();}
return x*f;
}
const LL maxn=401000,inf=1000000000000000LL,limit=20000,mod=1000000007;
bool cmin(LL& a,LL b){return a>b?a=b,true:false;}
bool cmax(LL& a,LL b){return a<b?a=b,true:false;}
namespace Mancher{
char ch[maxn];
LL p[maxn];
LL solve(char* s,LL n){
LL mx=0,id=0,m=0;
ch[m++]='&';ch[m++]='#';
up(i,0,n-1){
ch[m++]=s[i];
ch[m++]='#';
}
for(LL i=1;i<m;i++){
if(i<mx)p[i]=min(p[2*id-i],mx-i);
else p[i]=1;
while(ch[i+p[i]]==ch[i-p[i]])p[i]++;
if(i+p[i]>mx){
mx=i+p[i];
id=i;
}
}
LL ret=0;
for(LL i=1;i<m;i++)ret=(ret+p[i]/2)%mod;
return ret;
}
};
namespace FFT{
struct cp{
db x,y;
cp(db x=0,db y=0):x(x),y(y){}
cp operator+(const cp& b){return cp(x+b.x,y+b.y);}
cp operator-(const cp& b){return cp(x-b.x,y-b.y);}
cp operator*(const cp& b){return cp(x*b.x-y*b.y,x*b.y+y*b.x);}
}w[maxn],A[maxn],B[maxn];
inline void swap(cp& a,cp& b){cp t(a);a=b;b=t;}
LL R[maxn],H=1,L=1;
db pi=(acos(-1.0));
void FFT(cp* a,LL f){
for(LL i=0;i<L;i++)if(i<R[i])swap(a[i],a[R[i]]);
for(LL len=2;len<=L;len<<=1){
LL l=len>>1;
cp wn(cos(pi/l),f*sin(pi/l));
for(LL i=1;i<l;i++)w[i]=w[i-1]*wn;
for(LL st=0;st<L;st+=len){
for(LL k=0;k<l;k++)	{
cp x=a[st+k],y=w[k]*a[st+k+l];
a[st+k]=x+y,a[st+k+l]=x-y;
}
}
}
if(f==-1)up(i,0,L-1)a[i].x/=L;
}
void solve(LL* c,LL* d,LL n,LL m,LL* ch){
for(H=0,L=1;L<n+m-1;H++)L<<=1;
up(i,0,L-1)A[i].x=c[i],A[i].y=0;
up(i,0,L-1)B[i].x=d[i],B[i].y=0;
w[0].x=1;
up(i,0,L)R[i]=(R[i>>1]>>1)|((i&1)<<(H-1));
FFT(A,1);FFT(B,1);
for(LL i=0;i<L;i++)A[i]=A[i]*B[i];
FFT(A,-1);
for(LL i=0;i<L;i++)ch[i]=(LL)(A[i].x+0.5);
}
};
LL n;
char s[maxn],ch[maxn];
LL a[maxn],b[maxn],c[maxn];
LL mul(LL a,LL b){LL ans=1;for(;b;a=a*a%mod,b>>=1)if(b&1)ans=ans*a%mod;return ans;}
LL f[maxn];
int main(){
freopen(FILE".in","r",stdin);
freopen(FILE".out","w",stdout);
scanf("%s",s);
n=strlen(s);
up(i,0,n-1)ch[i]=s[i];
LL y=Mancher::solve(ch,n);
up(i,0,n-1)a[i]=s[i]=='a';
up(i,0,n-1)b[i]=s[i]=='b';
FFT::solve(a,a,n,n,c);
for(int i=0;i<n<<1;i++)f[i]+=c[i];
FFT::solve(b,b,n,n,c);
for(int i=0;i<n<<1;i++)f[i]+=c[i];
LL ans=0;
for(int i=0;i<n<<1;i++)
ans=(ans+mul(2,(f[i]+1)/2)+mod-1)%mod;
ans=(ans%mod-y+mod)%mod;
printf("%lld\n",ans);
return 0;
}


posted @ 2017-03-04 14:11  CHADLZX  阅读(94)  评论(0编辑  收藏  举报