ACM-ICPC 2018 南京赛区网络预赛Skr

题意：求本质不同的回文子串的和
题解：先构造pam，然后根据pam的原理（ch表示在该节点表示的回文串两侧加上该字符）对于每个节点维护一个表示该节点字符串的值，加起来即可

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define LL long long
//#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
template<typename T>
inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>
inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=2000000+10,maxn=1000000+10,inf=0x3f3f3f3f;

struct PAM{
    int ch[N][10],fail[N],cnt[N],num[N],len[N],s[N],le[N];
    ll val[N],ans[N];
    int last,n,p;
    int newnode(int w,int c)
    {
        for(int i=0;i<10;i++)ch[p][i] = 0;
        cnt[p] = num[p] = 0;
        len[p] = w;val[p]=c;
        return p++;
    }
    void init()
    {
        p = last = n = 0;
        newnode(0,0);
        newnode(-1,0);
        s[n] = -1;
        fail[0] = 1;
    }
    int getfail(int x)
    {
        while(s[n-len[x]-1] != s[n]) x = fail[x];
        return x;
    }
    void add(int c)
    {
        s[++n] = c;
        int cur = getfail(last);
        if(!ch[cur][c]){
            int now = newnode(len[cur]+2,c);
            fail[now] = ch[getfail(fail[cur])][c];
            ch[cur][c] = now;
            num[now] = num[fail[now]] + 1;
        }
        last = ch[cur][c];
        cnt[last]++;
    }
    void cal()
    {
        ll pp=0;
        for(int i=0;i<p;i++)
        {
            for(int j=0;j<10;j++)if(ch[i][j])
            {
                if(ans[i])ans[ch[i][j]]=(ans[i]*10%mod+val[ch[i][j]]+qp(10,1+le[i])*val[ch[i][j]]%mod)%mod,
                    le[ch[i][j]]=le[i]+2;
                else
                {
                    if(len[i]&1)ans[ch[i][j]]=val[ch[i][j]],le[ch[i][j]]=1;
                    else ans[ch[i][j]]=(val[ch[i][j]]+val[ch[i][j]]*10%mod)%mod,le[ch[i][j]]=2;
                }
            }
            pp+=ans[i];if(pp>=mod)pp-=mod;
        }
        printf("%lld
",pp);
    }
}pam;
char s[N];
int main()
{
    pam.init();
    scanf("%s",s);
    int n=strlen(s);
    for(int i=0;i<n;i++)pam.add(s[i]-'0');
    pam.cal();
    return 0;
}
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