题干：

yah has n strings <s1,⋯,sn>, and he generates a sequence P by two steps:

P=<s1​,⋯,sn​>
Replace each si​ with all prefixes of itself.
An example is:

the n strings are < aab,ab >
first, P =< aab,ab >
then, P =< a,aa,aab,a,ab >
There are 26 positive integers d1​,⋯,d26​ that denote the difficulty to identify lowercase English letter a,b,⋯,z for yah.

The difficulty to identify a string strstr is (i=1∏∣str∣dstri) mod  m

Now, yah wants to calculate: for each string si(i=1,⋯,n), the number of strings in P that is prefix of si and more difficult than si.

Input
The first line contains two integers n,mn,m(1≤n≤105,1≤m≤2×105).

The second line contains 26 positive d1,⋯,d26, 0^50≤di≤105.

The following n lines, each line contains a string, the ith lines contains si, the sum of length of all strings is not great than 2×105.

Output
The only line contains n integers, the ith​ integer denoting the number of strings in P that is prefix of si​ but more difficult to identify for yah.

题意：
给定n代表n个字符串，m代表模数，再给定26个字母的每个字母的权值d，再定义字符串s的值:前i个字符对应的子串的值dd[i]= dd[i-1]*(d[s[i]])  %  m 。

输入n个字符串，求对于每个字符串，大于该字符串 的 值 的 前缀的 个数（值dd[i]= dd[i-1]*(d[s[i]])  %  m ），主要求前缀的个数

解题报告：

map暴力搞一搞就能过、、但是要真是比赛场还是老老实实写hash吧、、万一T了呢。

其实也可以字典树，存前缀出现的次数。也不难想。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
map<string,PII> mp; 
int n;
int d[MAX];
ll m;
string s[MAX],tmp;
int main()
{
  ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
  cin>>n>>m;
  for(int i = 0; i<26; i++) cin>>d[i];
  for(int i = 1; i<=n; i++) {
    cin>>s[i];
    tmp = "";
    int len = s[i].length();
    ll tmpp = 1;
    for(int j = 0; j<len; j++) {
      tmp += s[i][j];
      tmpp *= d[s[i][j]-'a'];
      tmpp %= m;
      auto it = mp.find(tmp);
      if(it == mp.end()) mp[tmp] = pm(tmpp,1);
      else mp[tmp].SS ++;
    } 
  }
  for(int i = 1; i<=n; i++) {
    tmp = "";
    int ans = 0;
    int tar = mp[s[i]].FF;
    int len = s[i].length();
    for(int j = 0; j<len; j++) {
      tmp += s[i][j];
      if(mp[tmp].FF > tar) ans += mp[tmp].SS;
    }
    cout << ans ;
    if(i == n) cout << "\n";
    else cout << " ";
    //printf("%d%c",ans,i == n ?'\n' :' ');
  }
  return 0 ;
}