Algorithm
Problem Name: beecrowd | 2712
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/2712
Vehicular Restriction
Por , Silva A M 
Brazil
Timelimit: 1
The vehicle restriction of São Paulo city is a restriction on the circulation of motor vehicles in the city. Implemented since 1996 with the aim of improving environmental conditions by reducing the burden of pollutants in the atmosphere, it has consolidated itself as an instrument to reduce congestion in the main highways of the city, during peak times. In the boundary ways, traffic of trucks and automobiles that are within the restriction is not allowed. There is a scale that determines on which days of the week which vehicles can not travel. This scale is governed by the last digit of the vehicle nameplate, being:
- Monday, final digit of board 1 and 2
- Tuesday, final digit of plate 3 and 4
- Wednesday, digit end of plate 5 and 6
- Thursday, final digit of plate 7 and 8
- Friday, final digit of plate 9 and 0
The drivers who are caught violating the traffic restriction are charged with a fine and four points in the driver's license.
Input
The first entry line represents the number of N tests (1 <= N <1000) that should be considered. The other entries are string with maximum size S (1 <= S <= 100) representing each card to be analyzed, so that each card is on a single entry line. The expected format for a valid car plate in São Paulo is "AAA-9999", such that A is a valid character in [A-Z], and 9 is a valid numeric digit in [0-9].
Output
The set of valid values as output are: MONDAY, TUESDAY, WEDNESDAY, THURSDAY and FRIDAY, according to the predefined constraint table, and FAILURE if the board does not present the defined pattern.
| Input Samples | Output Samples |
|
3 |
TUESDAY |
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
#include <string.h>
void check(char str[])
{
int i, j, len, a=1, b=1, c=1, x;
len=strlen(str);
if (len < 8 || len > 8)
printf("FAILURE\n");
else
{
for (i = 0; i < 3; ++i)
{
if (str[i] >= 'A' && str[i] <= 'Z')
a=1;
else
{
a=0;
break;
}
}
if (str[3]!='-')
b=0;
for (i = 4; i < len; ++i)
{
if (str[i] >= '0' && str[i] <= '9')
c=1;
else
{
c=0;
break;
}
}
if (a==0 || b==0 || c==0)
printf("FAILURE\n");
else
{
x=str[len-1]-'0';
switch (x)
{
case 0:
case 9:
printf("FRIDAY\n");
break;
case 1:
case 2:
printf("MONDAY\n");
break;
case 3:
case 4:
printf("TUESDAY\n");
break;
case 5:
case 6:
printf("WEDNESDAY\n");
break;
case 7:
case 8:
printf("THURSDAY\n");
break;
}
}
}
}
int main(void)
{
int i, n;
char str[20];
scanf("%i", &n);
for (i = 0; i < n; ++i)
{
scanf("%s", str);
check(str);
}
return 0;
}
Input
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<vector>
#include<math.h>
#include<string>
#include<list>
using namespace std;
#define ll long long
#define input scanf
#define output printf
#define Loop while
#define echo cout
#define ret return
#define MAX 999999999999999999
#define MIN 0
#define PI 3.1415
vector < bool> a(32);
map<int,string> b;
void si2()
{
b[1]="MONDAY";
b[2]="MONDAY";
b[0]="FRIDAY";
b[9]="FRIDAY";
b[8]="THURSDAY";
b[7]="THURSDAY";
b[5]="WEDNESDAY";
b[6]="WEDNESDAY";
b[3]="TUESDAY";
b[4]="TUESDAY";
}
int main(int argc, char** argv) {
//freopen("c.txt","w",stdout);
si2();
int t;
cin >> t;
cin.ignore();
string str;
while(t--)
{
cin >> str;
if(str.length()==8)
{
bool f=false;
for(int i=0;i < 3; i++>
{
if(str[i] >= 65 && str[i] <= 90)
{
continue;
}
else
{
f=true;
break;
}
}
if(f)cout << "FAILURE\n";
else
{
if(str[3]!='-')
{
cout << "FAILURE\n";
}
else
{
f=false;
for(int i = 4;i < = 7; i++>
{
if(str[i] < 48 || str[i] > 57)
{
f=true;
break;
}
}
if(f)cout << "FAILURE\n";
else
cout << b[str[7]-48] << endl;
}
}
}
else
{
printf("FAILURE\n");
}
}
ret 0;
}
Input
Output
#3 Code Example with Java Programming
Code -
Java Programming
import java.util.Scanner;
import java.util.Locale;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
sc.useLocale(Locale.ENGLISH);
Locale.setDefault(new Locale("en", "US"));
while(sc.hasNext()) {
int n = Integer.parseInt(sc.nextLine());
//AAA-9999
for(int x = 0 ; x < n ; x++) {
String[] s = new String[8];
String placa = sc.nextLine();
if (placa.matches("[A-Z]{3}(-)\\d{4}")) {
s = placa.split("");
if (s[7].equals("1") || s[7].equals("2")) System.out.println("MONDAY");
if (s[7].equals("3") || s[7].equals("4")) System.out.println("TUESDAY");
if (s[7].equals("5") || s[7].equals("6")) System.out.println("WEDNESDAY");
if (s[7].equals("7") || s[7].equals("8")) System.out.println("THURSDAY");
if (s[7].equals("9") || s[7].equals("0")) System.out.println("FRIDAY");
}
else
System.out.println("FAILURE");
}
}
sc.close();
}
}
Input
Output
#4 Code Example with Javascript Programming
Code -
Javascript Programming
const { readFileSync } = require("node:fs")
const [N, ...input] = readFileSync("/dev/stdin", "utf8").split("\n", 1 + 1e3)
const vehicleLicensePlateFormatValidator = (plate = "") => /^[A-Z]{3}-[0-9]{4}$/.test(plate)
function main() {
const size = Number.parseInt(N, 10)
const output = Array.from(
{ length: size },
(_, index) => {
const plate = input[index]
if (vehicleLicensePlateFormatValidator(plate))
switch (plate.slice(-1)) {
case "1": case "2": return "MONDAY"
case "3": case "4": return "TUESDAY"
case "5": case "6": return "WEDNESDAY"
case "7": case "8": return "THURSDAY"
case "9": case "0": return "FRIDAY"
}
else return "FAILURE"
})
console.log(output.join("\n"))
}
main()
Input
Output
#5 Code Example with Python Programming
Code -
Python Programming
n = int(input())
while n:
n -= 1
placa = input().split('-')
if (len(placa) == 2) and (len(placa[0]) == 3) and (len(placa[1]) == 4) and (placa[0] == placa[0].upper()):
try:
check = int(placa[1])
r = int(placa[1][3])
if r > 8 or r == 0:
print('FRIDAY')
elif r > 6:
print('THURSDAY')
elif r > 4:
print('WEDNESDAY')
elif r > 2:
print('TUESDAY')
else:
print('MONDAY')
except:
print('FAILURE')
else:
print('FAILURE')
Input
Output