HDU 5785 Interesting

求出（1，i-1）中与i构成回文的位置的和L[i]，以及 i 与（i+1,n）中构成回文的位置和R[i]。

然后枚举每一对L[i]*R[i+1]，累加和就是答案。

计算L[i]和R[i]，可以利用manacher结果，o(n)处理得到。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-8;
void File()
{
    freopen("D:\in.txt","r",stdin);
    freopen("D:\out.txt","w",stdout);
}
inline int read()
{
    char c = getchar();  while(!isdigit(c)) c = getchar();
    int x = 0;
    while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }
    return x;
}

const int maxn=1000010;
char str[maxn],tmp[maxn<<1];
int Len[maxn<<1],n;
LL outp[maxn],out[maxn];
LL L[maxn],R[maxn];
LL mod=1000000007;
struct Seg { int l,r,zx; } seg[maxn<<1];
int cnt;

void init(char *st)
{
    int len=strlen(st); tmp[0]='@';
    for(int i=1;i<=2*len;i+=2) tmp[i]='#',tmp[i+1]=st[i/2];
    tmp[2*len+1]='#', tmp[2*len+2]='$', tmp[2*len+3]=0;
}

void Manacher(char *st)
{
    int mx=0,po=0,len=strlen(st);
    for(int i=1;i<=len;i++)
    {
        if(mx>i) Len[i]=min(mx-i,Len[2*po-i]);
        else Len[i]=1;
        while(st[i-Len[i]]==st[i+Len[i]]) Len[i]++;
        if(Len[i]+i>mx) mx=Len[i]+i, po=i;
    }
    Len[0]=1;
}

int main()
{
   // File();
    while(~scanf("%s",str))
    {

        n=strlen(str);
        init(str); Manacher(tmp);

        memset(outp,0,sizeof outp); memset(out,0,sizeof out);
        int LEN=strlen(tmp);

        cnt=0;
        for(int i=1;i<LEN-1;i++)
        {
            if(tmp[i]=='#'&&Len[i]==1) continue;
            int l=i-Len[i]+1,r=i+Len[i]-1;
            if(tmp[l]=='#') l++,r--; l=l/2,r=r/2;
            seg[cnt].l=l, seg[cnt].r=r, seg[cnt].zx=(l+r+1)/2, cnt++;
        }

       LL sum=0,sz=0;int pos=0;
       for(int i=1;i<=n;i++)
       {
           while(pos<cnt&&seg[pos].zx<=i)
           {
               sum=sum+seg[pos].l+seg[pos].r; sz++;
               out[seg[pos].r]-=1; outp[seg[pos].r]-=(seg[pos].l+seg[pos].r); pos++;
           }
           L[i]=sum-sz*i;
           sum=sum+outp[i]; sz=sz+out[i];
       }

       for(int i=0;i<cnt;i++) seg[i].zx=(seg[i].l+seg[i].r)/2;
       memset(outp,0,sizeof outp); memset(out,0,sizeof out);
       sum=0,sz=0,pos=cnt-1;

       for(int i=n;i>=1;i--)
       {
           while(pos>=0&&seg[pos].zx>=i)
           {
               sum=sum+seg[pos].l+seg[pos].r; sz++;
               out[seg[pos].l]-=1; outp[seg[pos].l]-=(seg[pos].l+seg[pos].r); pos--;
           }
           R[i]=sum-sz*i;
           sum=sum+outp[i]; sz=sz+out[i];
       }

       LL ans=0;
       for(int i=1;i<=n-1;i++) ans=(ans+(L[i]%mod)*(R[i+1]%mod)%mod)%mod;
       printf("%lld
",ans);
    }
    return 0;
}