F. Find / -type f -or -type d
题意
找到".eoj"结尾的"文件"（注意是".eoj"不是"eoj"）
思路
One more thing, on your file system, directory is only a logical concept. This means, a directory is created only when there is a file which relies on this directory is created and a directory cannot exist without files.
The root folder alone will not be included in this list.
目录不会脱离文件单独存在，意思就是如果一个字符串不是某个字符串的子串的话，那么这个字符串一定是文件，只要查看最后是否以".eoj"结尾即可，
但是要注意的是，文件并不一定不是某个字符串的子串，比如/a.eoj和/a.eojstr中/a.eoj也可能是文件
那么我们就要再次进行讨论，如果后续所有以该字符串为子串的字符串都是文件而不是目录的话，那么就可以判定该字符串为文件，然后查看最后是否以".eoj"结尾即可，
至于判断一个字符串是否是某个字符串的子串，我们可以先进行排序，得到的序列便是目录靠前的序列，然后用find函数逐个对比即可
代码
点击查看代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<cctype>
using namespace std;
#define X first
#define Y second
typedef pair<int,int> pii;
typedef long long LL;
const char nl = '\n';
const int N = 1e6+10;
const int M = 2e5+10;
int n,m;
vector<string> v;
void solve(){
cin >> m;
int ans = 0;
while(m -- ){
string s;
cin >> s;
v.push_back(s);
}
sort(v.begin(),v.end());
//for(auto s:v)cout << s << nl;
for(int i = 0; i < v.size() - 1; i ++ ){
if(v[i].size() <= 5)continue; //不满足长度
string t = v[i].substr(v[i].size() - 4,4);
if(t != ".eoj")continue;
if(v[i + 1].find(v[i]) == -1)ans ++;//i不是i+1的子串
else{//i是i+1的子串
int j = i + 1;
bool f = 0;
while(j <= v.size() - 1 && v[j].find(v[i]) != -1){
string tt = v[j];
for(int k = v[i].size() - 1; k <= tt.size() - 1; k ++ ){
if(tt[k] == '/'){
f = 1;
break;
}
}
j ++;
}
if(!f)ans ++;
}
}
string ss = v[v.size() - 1];
if(v[v.size() - 1].size() > 5){
string t = v[v.size() - 1].substr(v[v.size() - 1].size() - 4,4);
if(t == ".eoj"){
ans ++;
}
}
cout << ans;
}
int main(){
ios::sync_with_stdio(false);
cin.tie(0),cout.tie(0);
solve();
}