【HDU4622】Reincarnation

一眼似乎不可做，但发现\(strlen(x)\)很小，暴力\(O(n^2)\)预处理每个区间\((l,r)\)，查询时\(O(1)\)输出就好了

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
typedef int LL;
const LL maxn=2010;
inline LL Read(){
    LL x=0,f=1; char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-') f=-1; c=getchar();
    }
    while(c>='0'&&c<='9')
        x=(x<<3)+(x<<1)+c-'0',c=getchar();
    return x*f;
}
struct node{
    LL len,fail;
    LL son[26];
}t[maxn<<1];
LL n,m,T,nod,last;
LL ans[maxn][maxn];
char s[maxn];
inline void Insert(LL c){
    LL np=++nod,p=last;
    last=np;
    t[np].len=t[p].len+1;
    while(p&&!t[p].son[c]){
        t[p].son[c]=np,
        p=t[p].fail;
    }
    if(!p)
        t[np].fail=1;
    else{
        LL q=t[p].son[c];
        if(t[q].len==t[p].len+1)
            t[np].fail=q;
        else{
            LL nq=++nod;
            t[nq]=t[q];
            t[nq].len=t[p].len+1;
            t[np].fail=t[q].fail=nq;
            while(p&&t[p].son[c]==q){
                t[p].son[c]=nq,
                p=t[p].fail;
            }
        }
    }
}
int main(){
    T=Read();
    while(T--){
        scanf(" %s",s+1);
        LL Len=strlen(s+1);
        for(LL i=1;i<=Len;++i){
            nod=last=1;
            memset(t,0,sizeof(t));
            for(LL j=i;j<=Len;++j){
                Insert(s[j]-'a'),
                ans[i][j]=ans[i][j-1]+t[last].len-t[t[last].fail].len;
            }
        }
        m=Read();
        while(m--){
            LL l=Read(),r=Read();
            printf("%d\n",ans[l][r]);
        }
    }
    return 0;
}