題意:求本質不同的回文串(大整數)的數字和

由回文樹的性質可知貢獻只在首次進入某個新節點時產生

那麽只需由pos和len算出距離把左邊右邊刪掉再算好base重復\(O(n)\)次即可

位移那段寫的略微淩亂..

#include<bits/stdc++.h>
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define rrep(i,j,k) for(int i=j;i>=k;i--)
#define println(a) printf("%lld\n",(ll)(a))
#define printbk(a) printf("%lld ",(ll)(a))
typedef long long ll;
using namespace std;
const int MAXN = 2e6+11;
const ll oo = 0x3f3f3f3f3f3f3f3f;
const ll ooo= 0x3f3f3f3f;
const int MOD = 1000000007;
ll read(){
   ll x=0,f=1;register char ch=getchar();
   while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
   while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
   return x*f;
}
char str[MAXN];
ll _10[MAXN],inv10[MAXN];
ll sum[MAXN],rsum[MAXN];
ll ANS;
int llen;
typedef long long LL;
LL pow_mod(LL a, LL b, LL p){//a的b次方求余p 
    LL ret = 1;
    while(b){
        if(b & 1) ret = (ret * a) % p;
        a = (a * a) % p;
        b >>= 1;
    }
    return ret;
}
LL fermat(LL a, LL p){//費馬求a關於b的逆元 
        return pow_mod(a, p-2, p);
}
struct PT{
    char s[MAXN];
    int last,cur,tot;
    int son[MAXN][10];
    int fail[MAXN],len[MAXN];
    void init(){
        s[0]=-1;
        last=cur=0;
        tot=1;
        rep(i,0,9) son[0][i]=son[1][i]=0;
        len[0]=0,len[1]=-1;
        fail[0]=1;fail[1]=0;
        
    }
    int node(int l){
        ++tot;
        rep(i,0,9) son[tot][i]=0;
        fail[tot]=0;
        len[tot]=l;
        return tot;
    }
    int getfail(int x){
        while(s[cur-len[x]-1]^s[cur]) x=fail[x];
        return x;
    }
    void add(int pos){
        s[++cur]=str[pos];
        int t=getfail(last);
        int c=str[pos]-‘0‘;
        if(son[t][c]==0){
            int o=node(len[t]+2);
            fail[o]=son[getfail(fail[t])][c];
            son[t][c]=o;
            ll lo=((sum[llen]-rsum[pos-(len[t]+2)+1])%MOD+MOD)%MOD;
            ll hi=rsum[pos+1];
            ll all=lo+hi; if(all>=MOD) all%=MOD;
            ll t=((sum[llen]-all)%MOD+MOD)%MOD;
            if(t>=MOD) t%=MOD; 
            if(llen-pos+1-1>0)ANS=ANS+t*inv10[llen-pos+1-1]%MOD;
            else ANS=ANS+t;
            if(ANS>=MOD) ANS%=MOD;
        }
        last=son[t][c];
    }
}pt;
int main(){
    _10[0]=1;
    rep(i,1,MAXN-1){
        _10[i]=_10[i-1]*10;
        if(_10[i]>=MOD) _10[i]%=MOD;
        inv10[i]=fermat(_10[i],MOD);
    }
    while(~scanf("%s",str+1)){
        memset(sum,0,sizeof sum);
        memset(rsum,0,sizeof rsum);
        ANS=0;
        sum[0]=0;
        pt.init();
        llen=strlen(str+1);
        rep(i,1,llen){
            sum[i]=sum[i-1]*10+(int)(str[i]-‘0‘);
            if(sum[i]>=MOD) sum[i]%=MOD;
        }
        rsum[llen+1]=0;
        rrep(i,llen,1){
            rsum[i]=rsum[i+1]+(int)(str[i]-‘0‘)*_10[llen-i]%MOD;
            if(rsum[i]>=MOD) rsum[i]%=MOD;
        }
        ANS=0;
        rep(i,1,llen) pt.add(i);
        println(ANS%MOD);
    }
    return 0;
}
 

2018南京網絡賽 - Skr 回文樹