Algorithm
Problem Name: 2 AD-HOC - beecrowd | 1769
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/1769
SSN 1
By Alexandre Campos, UNIUBE 
Brazil
Timelimit: 1
In Brazil, the equivalent to Social Security Number is usually called CPF. It is composed by 11 digits and the two lasts 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
a1a2a3.a4a5a6.a7a8a9-b1b2
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.
Analogous, 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.
Given a CPF number, you have to tell whether it is valid or not.
Input
The input is composed by an unknown number of CPF numbers, not more than 10000 cases. Each line has a CPF in the form
d1d2d3.d4d5d6.d7d8d9-d1d2
Output
If the given CPF is valid, print "CPF valido". Otherwise, print "CPF invalido".
| Input Sample | Output Sample |
|
048.856.829-63 |
CPF invalido |
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
#include <ctype.h>
void transformaInt(char *, short vet[]);
void retiraPonto(char *, char *);
int main (void)
{
char cpf[15], cpfCpy[15];
short numCPF[15], numB1 = 0, numB2 = 0;
unsigned char numero, i = 0, j , k;
while ((scanf(" %s", cpf) != EOF ))
{
//Retirar caracteres especiais;
retiraPonto(cpf, cpfCpy);
//Converter caractere para inteiro;
transformaInt(cpfCpy, numCPF);
for (i = 0, j = 1, k = 9; i < 9; i++, j++, k--)
{
numB1 += numCPF[i] * j;
numB2 += numCPF[i] * k;
}
numB1 = numB1 % 11;
if (numB1 % 11 == 10)
numB1 = 0;
numB2 = numB2 % 11;
if (numB2 % 11 == 10)
numB2 = 0;
if (numB1 == numCPF[9] && numB2 == numCPF[10])
printf("CPF valido\n");
else
printf("CPF invalido\n");
numB1 = 0;
numB2 = 0;
}
}
void transformaInt(char *str, short vet[])
{
unsigned short i;
for (i = 0; i < 11; i++)
vet[i] = (int)str[i] - 48;
}
void retiraPonto(char *str, char *strCpy)
{
short i = 0, j = 0;
while (i < 14)
{
if (isdigit(str[i]))
strCpy[j++] = str[i];
i++;
}
strCpy[j] = '\0';
}
Input
733.184.680-96
227.518.471-08
092.844.842-86
098.447.895-55
Output
CPF valido
CPF invalido
CPF valido
CPF invalido
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
using namespace std;
void cpf(char *a, char *b) {
while (*a) {
if ('0' < = *a and *a <= '9')
*b++ = *a;
++a;
}
*b = '\0';
}
short cti(char c) {
return c - 48;
}
int main(void) {
int s, i, b1, b2;
char e[15], c[12];
while (cin >> e) {
cpf(e, 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;
if (b1 == cti(c[9]) and b2 == cti(c[10]))
cout << "CPF valido";
else
cout << "CPF invalido";
cout << endl;
}
return 0;
}
Input
733.184.680-96
227.518.471-08
092.844.842-86
098.447.895-55
Output
CPF valido
CPF invalido
CPF valido
CPF invalido
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const { readFileSync } = require("fs")
const input = readFileSync("/dev/stdin", "utf8").split("\n")
class CPF {
/** @param {string} cpf */
static validate(cpf) {
const arr = cpf
.match(/\d/g)
.slice(0, 11)
.map((code) => Number.parseInt(code, 10))
let res01 = 0
let res02 = 0
for (let index = 0; index < 9; index++) {
res01 += (index + 1) * arr[index]
res02 += (9 - index) * arr[index]
}
return (res01 % 11) % 10 == arr[9] && (res02 % 11) % 10 == arr[10]
}
}
function main() {
const responses = []
for (const line of input)
if (line == "") break
else responses.push(CPF.validate(line) ? "CPF valido" : "CPF invalido")
console.log(responses.join("\n"))
}
main()
Input
733.184.680-96
227.518.471-08
092.844.842-86
098.447.895-55
Output
CPF valido
CPF invalido
CPF valido
CPF invalido
#4 Code Example with Python Programming
Code -
Python Programming
while True:
try:
e = str(input()).strip()
c = e.replace('.', '').replace('-', '')
s = 0
for i in range(9):
s += int(c[i]) * (i + 1)
b1 = s % 11
b1 = 0 if b1 == 10 else b1
s = 0
if b1 == int(e[-2]):
for i in range(9):
s += int(c[i]) * (9 - i)
b2 = s % 11
b2 = 0 if b2 == 10 else b2
if b2 == int(e[-1]):
print('CPF valido')
else: print('CPF invalido')
else: print('CPF invalido')
except EOFError: break
Input
733.184.680-96
227.518.471-08
092.844.842-86
098.447.895-55
Output
CPF valido
CPF invalido
CPF valido
CPF invalido