URI Online Judge Solution 1786 | SSN 2

URI Online Judge | 1786

SSN 2

By Alexandre Campos, UNIUBE  Brazil

Timelimit: 1

You are about to write a program to predict a CPF, which, in Brazil, is equivalent to Social Security Number. It is composed by 11 digits and the lasts two (verification digits) are function of the nine previous. In this way, if a person informs a CPF, by mistake or on purpose, it is possible to find out. Let us introduce some notation. Let a CPF be

a1 a2 a3 . a4 a5 a6 . a7 a8 a9 - b1 b2

To get b1, one can multiply a1 by 1, a2 by 2, a3 by 3, so on, up to a9 by 9 and sum these results. Then, b1 is the remaining of this number when divided by 11, or 0 in case the remaining is 10.

Analogously, to get b2, one can multiply a1 by 9, a2 by 8, a3 by 7, so on, up to a9 by 1 and sum these results. Then, b2 is the remaining of this number when divided by 11, or 0 in case the remaining is 10.

Input

The input is composed by an unknown number of sequences in the form:

a1a2a3a4a5a6a7a8a9

Each sequence represents the 9 firsts digits of a CPF.

Output

For each sequence, you have to print the input sequence and the verification digits formated as

a1a2a3.a4a5a6.a7a8a9-b1b2

Input Sample	Output Sample
000000000 111111111 354122447 569961340 169992467	000.000.000-00 111.111.111-11 354.122.447-93 569.961.340-48 169.992.467-85

URI Online Judge Solution 1786 | SSN 2 in C++

#include <iostream>

#include <cstring>

 

#define for(i, n) for(int i = 0; i < (n); ++i)

    

using namespace std;

    

int main(int argc, char const *argv[])

{

    int b1, b2, l, r;

    string s;

    

    while(getline(cin, s))

    {

        l = 1, r = 9, b1 = b2 = 0;

 

        for(i, 9)

        {

            b1 += (l * (s[i] - '0'));

            b2 += (r * (s[i] - '0'));

            l++; r--;

        }

 

        if(b1 % 11 == 10) b1 = 0;

        else b1 %= 11;

        if(b2 % 11 == 10) b2 = 0;

        else b2 %= 11;

 

        cout << s[0] << s[1] << s[2] << "." << s[3] << s[4] << s[5] << "." << s[6] << s[7] << s[8] << "-" << b1 << b2 << endl;

    }

    

    return 0;

}