HDU

　　题意：给你一个字符串，给你Q次询问，每一次问你从l-r里有多少个回文串。

　　思路：len很小，所以直接遍历区间求就好了。

/*  gyt
       Live up to every day            */
#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<cstring>
#include<queue>
#include<set>
#include<string>
#include<map>
#include <time.h>
#define PI acos(-1)
using namespace std;
typedef long long ll;
typedef double db;
const int maxn = 1000+10;
const int sigma=26;
const ll mod = 1000000007;
const int INF = 0x3f3f3f;
const db eps = 1e-9;
struct ptree{
    char s[maxn];
    int next[maxn][sigma], fail[maxn], cnt[maxn], len[maxn];
    int last, n, p;
    ll res;
    inline int newnode(int l) {
        cnt[p]=0;
        len[p]=l;
        return p++;
    }
    inline void init() {
        n=0, p=0, last=0;
        memset(next, 0, sizeof(next));
        memset(cnt, 0, sizeof(cnt));
        memset(len, 0, sizeof(len));
        memset(fail, 0, sizeof(fail));
        newnode(0), newnode(-1);
        s[n]=-1;
        fail[0]=1;
        //cout<<n<<" "<<p<<" "<<last<<endl;
    }
    inline int FL(int x) {
        while(s[n-len[x]-1]!=s[n])  x=fail[x];
        return x;
    }
    void add(char c) {
        c-='a';
        s[++n]=c;
        int cur=FL(last);
        if (!next[cur][c]) {
            int now=newnode(len[cur]+2);
            fail[now]=next[FL(fail[cur])][c];
            next[cur][c]=now;
        }
        last=next[cur][c];
        ++cnt[last];
    }
    inline ll countt() {
        ll pk=0;
        for (int i=p-1; ~i; --i) {
            cnt[fail[i]]+=cnt[i];
            pk++;
        }
        return pk;
    }
}p;
char s[maxn];
int ans[maxn][maxn];
void solve(){
    memset(ans, 0, sizeof(ans));
    scanf("%s",s);
    int len=strlen(s);
    for (int i=0; i<len; i++) {
        p.init();
        for (int j=i; j<len; j++) {
            p.add(s[j]);
            ans[i][j] = p.p-2;
        }
    }
    int q;  scanf("%d", &q);
    while(q--) {
        int a, b;  scanf("%d%d", &a, &b);
        printf("%d
", ans[a-1][b-1]);
    }
}
int main() {
    int t = 1;
   // freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    scanf("%d", &t);
    while(t--)
        solve();
    return 0;
}