只有24分

#include <bits/stdc++.h>
using namespace std;

const int base=31,mod=1e7+7;
int n,ans,p[20],cnt[mod+10];//p[i]=base^i,cnt[i]=哈希值为i的字符串出现次数
map<int,int>used;//used[i]记录哈希值为i的字符串是否使用过
string s[20010];
int hsh[20010][25];//hsh[i][j]=第i个密码前j个字符的哈希值

int main(){
    cin>>n;
    for (int i = 1; i <= n; i++) cin>>s[i];
    //初始化
    p[0] = 1;
    for (int i = 1; i <= n; i++){
        p[i] = 1ll * p[i-1] * base % mod; 
        for (int j = 1; j <= s[i].size(); j++)
            hsh[i][j] = (1ll * hsh[i][j-1] * base % mod + s[i][j-1] - 'a' + 1) % mod;
    }
    for (int i = 1; i <= n; i++){//枚举每个字符串
        used.clear();//换了字符串要清除
        for(int j = 1; j <= s[i].size(); j++)//枚举左端点
            for (int k = 1; k <= s[i].size(); k++){//枚举长度
                int tip = (hsh[i][j+k-1] - 1ll * hsh[i][j-1] * p[k] % mod + mod) % mod;
                if(used[tip] == false){//一个字符串里同样的子串只能贡献一次次数
                    used[tip] = true;
                    cnt[tip]++;
                }
            }
    }
    //枚举每个字符串，看是否为其他字符串的子串
    for (int i = 1; i <= n; i++)
        if(cnt[hsh[i][s[i].size()]] > 0)
            ans += cnt[hsh[i][s[i].size()]] - 1;//减掉自己
    cout<<ans;
}