hdoj5785

题意：略

先用题解的办法，manacher,然后tag,add数组。但是比较难办的是manacher加了新的字符。这样的话cntL和cntR不是实际的值，但是没关系，原本的字符都在奇数位置，这样cntL[i]就等于(add[i]-tag[i])/2就是真实值，具体来说不好看，我看了3个小时才明白。比如

$#a#a#a#

12345678(下标)

01234321(radius)

那么第一个a,i=3时cntL+3,i=4时cntL+5,i=5时cntL+7。实际上是+1，+2,+3的。而tag表示它被加过几次。因此1+2+3=(3+5+7-3)/2；除法用逆元。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <cstring>
#include <map>
#include <queue>
#include <set>
#include <cassert>
#include <stack>
#include <bitset>
#define mkp make_pair
using namespace std;
const double EPS=1e-;
typedef long long lon;
const lon SZ=,SSZ=*SZ,INF=0x7FFFFFFF,mod=;
char src[SSZ],ch[SSZ];
lon slen,clen,rd[SSZ];
lon cntr[SSZ],cntl[SSZ];
lon add[SSZ],tag[SSZ],inv;

void init()
{
    slen=strlen(src+);
    clen=;
    ch[++clen]='$';
    ch[++clen]='#';
    for(lon i=;i<=slen;++i)
    {
        ch[++clen]=src[i];
        ch[++clen]='#';
    }
}

lon ext(lon ct,lon pos,lon &maxr,lon &maxid)
{
    lon res=;
    for(lon i=pos;i<=clen;++i)
    {
        lon ll=ct-(i-ct);
        if(ch[ll]==ch[i])++res;
        else break;
    }
    return res;
}

void manacher()
{
    lon maxr=,maxid=;
    for(lon i=;i<=clen;++i)
    {
        if(i<=maxr)
        {
            lon rf1=maxid-(i-maxid);
            lon rf2=maxid-(maxr-maxid);
            lon ll=(rf1-rd[rf1]+);
            if(ll>rf2)rd[i]=rd[rf1];
            else rd[i]=(maxr-i+)+ext(i,maxr+,maxr,maxid);
        }
        else
        {
            rd[i]=ext(i,i,maxr,maxid);
        }
        if(i+rd[i]->maxr)
        {
            maxr=i+rd[i]-;
            maxid=i;
        }
    }
}

void work()
{
    manacher();
    for(lon i=;i<=clen;++i)
    {
        lon ll=i-rd[i]+,rr=i+rd[i]-;
        ++tag[ll],add[ll]+=(i+rd[i]-)+;
        --tag[i+],add[i+]-=i;
    }
    for(lon i=;i<=clen;++i)
    {
        //cout<<tag[i]<<" "<<add[i]<<endl;
        tag[i]+=tag[i-];
        add[i]+=add[i-]-tag[i];
        add[i]%=mod;
        cntr[i]=(add[i]-tag[i])*inv%mod;
        //cout<<cntr[i]<<endl;
    }
    for(lon i=;i<=clen+;++i)
    {
        tag[i]=add[i]=;
    }
    for(lon i=;i<=clen;++i)
    {
        lon rr=i+rd[i]-,ll=i-rd[i]+;
        ++tag[rr],add[rr]+=(i-rd[i]+)-;
        --tag[i-],add[i-]-=i;
    }
    for(lon i=clen;i>=;--i)
    {
        tag[i]+=tag[i+];
        add[i]+=add[i+]+tag[i];
        add[i]%=mod;
        cntl[i]=(add[i]-tag[i])*inv%mod;
    }
    for(lon i=;i<=clen+;++i)
    {
        tag[i]=add[i]=;
    }
    lon res=;
    for(lon i=;i<clen;i+=)
    {
        //cout<<cntl[i]<<" "<<cntr[i+1]<<endl;
        res+=cntl[i]*cntr[i+];
        res%=mod;
    }
    res=(res+mod)%mod;
    cout<<res<<endl;
    for(lon i=;i<=clen+;++i)
    {
        cntl[i]=cntr[i]=rd[i]=;
    }
}

void ex_gcd(lon a,lon b,lon &x,lon &y,lon &d)
{
    if(b==)
    {
        d=a,x=,y=;
    }
    else
    {
        ex_gcd(b,a%b,y,x,d);
        y-=(a/b)*x;
    }
}

int main()
{
    //std::ios::sync_with_stdio(0);
    //freopen("d:\\1.txt","r",stdin);
    lon x,y,d;
    ex_gcd(,mod,x,y,d);
    inv=x;
    lon casenum;
    //cin>>casenum;
    //cout<<casenum<<endl;
    //for(lon time=1;time<=casenum;++time)
    for(int time=;scanf(" %s",src+)!=EOF;++time)
    {
        init();
        work();
    }
    return ;
}