Algorithm

A. Jabber ID
time limit per test
0.5 second
memory limit per test
256 megabytes
Problem link- https://codeforces.com/problemset/problem/21/A
input
standard input
output
standard output

Jabber ID on the national Berland service «Babber» has a form <username>@<hostname>[/resource], where

  • <username> — is a sequence of Latin letters (lowercase or uppercase), digits or underscores characters «_», the length of <username> is between 1 and 16, inclusive.
  • <hostname> — is a sequence of word separated by periods (characters «.»), where each word should contain only characters allowed for <username>, the length of each word is between 1 and 16, inclusive. The length of <hostname> is between 1 and 32, inclusive.
  • <resource> — is a sequence of Latin letters (lowercase or uppercase), digits or underscores characters «_», the length of <resource> is between 1 and 16, inclusive.

The content of square brackets is optional — it can be present or can be absent.

There are the samples of correct Jabber IDs: mike@codeforces.com, 007@en.codeforces.com/contest.

Your task is to write program which checks if given string is a correct Jabber ID.

Input

The input contains of a single line. The line has the length between 1 and 100 characters, inclusive. Each characters has ASCII-code between 33 and 127, inclusive.

Output

Print YES or NO.

Examples
input
Copy
mike@codeforces.com
output
Copy
YES
input
Copy
john.smith@codeforces.ru/contest.icpc/12
output
Copy
NO

 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

int const N = 1e5 + 1;
long long const M = 1e5 * 1e6 + 1;
int n, m, parent[N];
long long cost[N];
bool vis[N];
vector<vector<pair<int, int> > > g;
vector<int> res;
priority_queue<pair<long long, int> > qu;

void Dijkstra(int src) {
    while(!qu.empty()) qu.pop();
    for(int i = 0; i < n; i++) cost[i] = M;
    cost[src] = 0;
    qu.push(make_pair(0, src));
    
    while(!qu.empty()) {
        int v = qu.top().second;
        long long c = -qu.top().first;
        
        qu.pop();
        if(vis[v]) continue;
        vis[v] = true;
        
        for(int i = 0; i < g[v].size(); i++)
            if(g[v][i].second + c < cost[g[v][i].first]) {
                parent[g[v][i].first] = v;
                cost[g[v][i].first] = g[v][i].second + c;
                qu.push(make_pair(-(g[v][i].second + c), g[v][i].first));
            }
    }
}

int main() {
    int u, v, c;
    
    cin >> n >> m;
    g.resize(n);
    
    for(int i = 0; i < m; i++) {
        cin >> u >> v >> c;
        u--; v--;
        g[u].push_back(make_pair(v, c));
        g[v].push_back(make_pair(u, c));
    }
    
    Dijkstra(0);
    
    if(cost[n - 1] == M) cout << -1 << endl;
    else {
        v = n - 1;
        
        while (v != 0) {
            res.push_back(v + 1);
            v = parent[v];
        }
        
        res.push_back(1);

        for (int i = res.size() - 1; i >= 0; i--) cout << res[i] << ' ';
        cout << endl;
    }
    
    return 0;
}
Copy The Code & Try With Live Editor

Input

x
–
+
cmd
mike@codeforces.com

Output

x
–
+
cmd
YES
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Demonstration

Codeforcess Solution 21A. Jabber IDC,C++, Java, Js and Python ,21A. Jabber,Codeforcess Solution

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