AC记录https://hydro.ac/d/atcoder/record/66bee0bb3f788cb34ad205e1

分析

用count函数就行

#include <bits/stdc++.h>
#include <algorithm>

#define endl '\n';


const int N = 1e6 + 10;
const int M = 2e6 + 10;
const int P = 998244353; 
const int Base = 3221225477;
const int INF = 0x3f3f3f3f3f3f3f3f; 

int mod(int x)
{
   return x % P;
}
using namespace std;
string s;
signed main()
{
	cin >> s;
	for(int i = 0;i < s.size();i++)
	{
		int t  = count(s.begin(),s.end(),s[i]);
		if(t > 1)
		{
			cout << "no" << endl;
			return 0;
		}
	}
	cout << "yes" << endl;
	return 0;
}