Algorithm
Problem Name: 2 AD-HOC - beecrowd | 1786
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/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 |
000.000.000-00 |
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
#include <string.h>
void makeInt(char *str, short vet[]);
int main (void)
{
char cpf[10];
short numCPF[9];
unsigned short digitoA, digitoB;
unsigned i, j, k;
while (scanf(" %s", cpf) != EOF)
{
makeInt(cpf, numCPF);
digitoA = 0;
digitoB = 0;
for (i = 0, j = 1, k = 9; i < 9; i++, j++, k--)
{
digitoA += numCPF[i] * j;
digitoB += numCPF[i] * k;
}
digitoA = digitoA % 11;
if (digitoA == 10)
digitoA = 0;
digitoB = digitoB % 11;
if (digitoB == 10)
digitoB = 0;
for (i = 0; i < 3; i++)
printf("%hu", cpf[i] - 48);
printf(".");
for (i = 3; i < 6; i++)
printf("%hu", cpf[i] - 48);
printf(".");
for (i = 6; i < 9; i++)
printf("%hu", cpf[i] - 48);
printf("-");
printf("%hu%hu", digitoA, digitoB);
printf("\n");
}
}
void makeInt(char *str, short vet[])
{
unsigned short i;
for (i = 0; i < 9; i++)
vet[i] = (int)str[i] - 48;
}
Input
111111111
354122447
569961340
169992467
Output
111.111.111-11
354.122.447-93
569.961.340-48
169.992.467-85
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
using namespace std;
short cti(char c) { return c - 48; }
int main(void) {
int s, i, b1, b2;
char c[10];
while (cin >> c) {
s = 0;
for (i = 0; i < 9; ++i)
s += cti(c[i]) * (i+1);
b1 = s % 11;
b1 = (b1 != 10) ? b1 : 0;
s = 0;
for (i = 0; i < 9; ++i)
s += cti(c[i]) * (9-i);
b2 = s % 11;
b2 = (b2 != 10) ? b2 : 0;
cout << c[0] << c[1] << c[2] << ".";
cout << c[3] << c[4] << c[5] << ".";
cout << c[6] << c[7] << c[8] << "-";
cout << b1 << b2 << endl;
}
return 0;
}
Input
111111111
354122447
569961340
169992467
Output
111.111.111-11
354.122.447-93
569.961.340-48
169.992.467-85
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const { readFileSync } = require("fs")
const input = readFileSync("/dev/stdin", "utf8").split("\n")
class CPF {
/** @param {number[]} ints */
static generateFromIntegers(ints) {
const digits = new Uint8Array(11)
let res01 = 0
let res02 = 0
for (let index = 0; index < 9; index++) {
digits.set([ints[index] || 0], index)
res01 += (index + 1) * digits[index]
res02 += (9 - index) * digits[index]
}
digits[9] = (res01 % 11) % 10
digits[10] = (res02 % 11) % 10
return digits.join("")
}
/** @param {string} cpf*/
static format(cpf) {
return cpf.replace(/(\d{3})(\d{3})(\d{3})(\d{2})/, "$1.$2.$3-$4")
}
}
function main() {
const responses = []
for (const line of input) {
if (line === "") break // EOFile
const generated = CPF.generateFromIntegers([...line].map(Number))
const formatted = CPF.format(generated)
responses.push(formatted)
}
console.log(responses.join("\n"))
}
main()
Input
111111111
354122447
569961340
169992467
Output
111.111.111-11
354.122.447-93
569.961.340-48
169.992.467-85
#4 Code Example with Python Programming
Code -
Python Programming
while True:
try:
e = str(input()).strip().strip()
s = 0
for i in range(9):
s += int(e[i]) * (i + 1)
b1 = s % 11
b1 = 0 if b1 == 10 else b1
s = 0
for i in range(9):
s += int(e[i]) * (9 - i)
b2 = s % 11
b2 = 0 if b2 == 10 else b2
print('{}{}{}.{}{}{}.{}{}{}-{}{}'.format(e[0], e[1], e[2], e[3], e[4], e[5], e[6], e[7], e[8], b1, b2))
except EOFError: break
Input
111111111
354122447
569961340
169992467
Output
111.111.111-11
354.122.447-93
569.961.340-48
169.992.467-85