梁成爱玩绝区零

对于一个字符串的子串最多有n*(n+1)/2，这题的数据只有500，最多只有250000个子串，可以考虑用map存下所以子串的个数(由于map会对字符串自动排序，可能会很慢，所以字符串的题还是用 unordered_map吧)，暴力判断即可，时间复杂度O(1)~O(n)，代码很短

#include<bits/stdc++.h>
using namespace std;
int const N=1e5+10;
unordered_map<string,int> mp;
int main()
{
	int n;
	cin>>n;
	string s1;
	cin>>s1;
	for(int i=0;i<n;i++)
	{
		string s="";
		for(int j=i;j<n;j++)
		{
			s+=s1[j];
			mp[s]++;
		}
	}
	int m,ans=0;
	cin>>m;
	for(int i=1;i<=m;i++)
	{
		string s;
		cin>>s;
		ans+=mp[s];
	}
	cout<<ans<<endl;
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
const int N = 5e2+5;
int n;
char ch[N];
int q;
int main()
{
	int ans=0;
	cin>>n;
	cin>>ch+1;
	cin>>q;
	while(q--)
	{
		string s;
		cin>>s;
		int len=s.length();
		for(int i=1;i<=n-len+1;i++)
		{
			bool flag=0;
			for(int j=0;j<=len-1;j++)
			{
				if(ch[i+j]!=s[j])
				{
					flag=1;
					break;
				}
			}
			if(!flag) ans++;
		}
	} 
	cout<<ans;
	return 0;
}

n=int(input())
s=input().strip('\n')
m=int(input())
ss=[""]*m
ans=0
for i in range(m):
    ss[i]=input().strip('\n')
for i in range(len(s)):
    for j in range(i,len(s)):
        if s[i:j+1] in ss:
            ans+=1
print(ans)

#include<bits/stdc++.h>
template<class T>inline void read(T &x){x=0;char c=getchar();bool f=0;for(;!isdigit(c);c=getchar())if(c=='-')f=1;for(;isdigit(c);c=getchar())x=(x<<3)+(x<<1)+c-'0';x=f?(~(x-1)):x;}
template<class T>inline void write(T x){if(x<0)return putchar('-'),write(-x),void();if(x>=10)write(x/10);putchar(x%10+'0');return;}
template<class T>inline void print(T x){write(x);putchar('\n');}
template<typename type,typename ...T>inline void read(type &x,T&...y){read(x),read(y...);}
using namespace std;
const int N=505;
int n,q;
string s,t[N];
inline int check(int pos){
	int ans=0;
	for(int i=1;i<=q;++i){
		int len=t[i].length()-1;
		bool same=1;
		if(pos+len-1>n)same=0;
		else{
			for(int j=1;pos+j-1<=n&&j<=len;++j){
				if(t[i][j]!=s[pos+j-1]){
					same=0;
					break;
				}
			}
		}
		if(same==1){
			ans++;
		}
	}
	return ans;
}
int result;
signed main(){
	read(n);
	cin>>s;
	s=" "+s;
	read(q);
	for(int i=1;i<=q;++i){
		cin>>t[i];
		t[i]=" "+t[i];
	}
	for(int i=1;i<=n;++i){
		result+=check(i);
	}
	print(result);
	return 0;
}