URI - BEECROWD - BEE 1050 - DDD Solution in C,C++,Python | URI - BEECROWD - BEE Solution 1050 - DDD

URI - BEECROWD - BEE 1050 - DDD Solution in C,C++,Python | URI - BEECROWD - BEE Online Judge Solution 1050 - DDD

Read an integer number that is the code number for phone dialing. Then, print the destination according to the following table:

If the input number isn’t found in the above table, the output must be:
DDD nao cadastrado
That means “DDD not found” in Portuguese language.

Input

The input consists in a unique integer number.

Output

Print the city name corresponding to the input DDD. Print DDD nao cadastrado if doesn't exist corresponding DDD to the typed number.

Input Sample	Output Sample
11	Sao Paulo

URI Online Judge Solution 1050 - DDD | URI 1050 Solution in C,C++,Python :

                                        Demonstration:

It's a not a hard problem.You just have to check the conditions by using if-else or switch statement to find the combination of the table given.Then you have to print those values according to the problem's statement.

N.B: Don't copy paste the code as same. Just try to understand it and try yourself. It would be better for you.

  URI Problem 1050 Solution in C :   

URI Online Judge 1050 Solve  in C :

#include <stdio.h>
int main()
{
   int N;
    scanf("%d",&N);
    if(N==61)
        printf("Brasilia\n");
    else if(N==71)
        printf("Salvador\n");
    else if(N==11)
        printf("Sao Paulo\n");
    else if(N==21)
        printf("Rio de Janeiro\n");
    else if(N==32)
        printf("Juiz de Fora\n");
    else if(N==19)
        printf("Campinas\n");
    else if(N==27)
        printf("Vitoria\n");
    else if(N==31)
        printf("Bela Horizonte\n");
    else
        printf("DDD nao cadastrado\n");
    return 0;
}

  URI Problem 1050 Solution in C ++:

URI Online Judge 1050 Solve  in C++ :

#include <bits/stdc++.h>
using namespace std;


int main(){
    int ddd;
    cin >> ddd;
    switch(ddd){
              case 11:
                   cout << "Sao Paulo"<<endl;
                   break;
              case 19:
                   cout << "Campinas"<<endl;
                   break;
              case 21:
                   cout << "Rio de Janeiro"<<endl;
                   break;
              case 27:
                   cout << "Vitoria"<<endl;
                   break;
              case 31:
                   cout << "Belo Horizonte"<<endl;
                   break;
              case 32:
                   cout << "Juiz de Fora"<<endl;
                   break;
              case 61:
                   cout << "Brasilia"<<endl;
                   break;
              case 71:
                   cout << "Salvador"<<endl;
                   break;
              default:
                      cout << "DDD nao cadastrado"<<endl;
                      break;

    }
    
return 0;
}

  URI Problem 1050 Solution in Python:

URI Online Judge 1050 Solve  in Python  :

a=int(input())
if a==61:
    print("Brasilia")
elif a==71:
    print("Salvador")
elif a==11:
    print("Sao Paulo")
elif a==21:
    print("Rio de Janeiro")
elif a==32:
    print("Juiz de Fora")
elif a==19:
    print("Campinas")
elif a==27:
    print("Vitoria")
elif a==31:
    print("Belo Horizonte")
else:
    print("DDD nao cadastrado")

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