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

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

## 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

### 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)
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