program to add one to each digit of a number - c

Question

As new to competitive programming, I was solving this practice question. The goal is to write a program to display numbers whose digits are 1 greater than the corresponding digits of the entered number. So if the number input is 12345 then the output number should be 23456. I have figured out how to separate each number and add them, but I was unable able to take a number of test cases in the following program.
The question is as follows
Input
First line of input will contain a number N = number of test cases. Next N lines will contain number n as test case where 1<=n<=99999.
Output
For each input case, add one to each digit of n, and print the new number.
As a beginner in competitive programming would be helpful if you give some tips to optimize the code.
here is the code that I have written.
#include<stdio.h>
void main()
{
int n, t, sum = 0;
scanf("%d", &t);
int a[t];
for (int j = 0; j < t; j++)
{
for (int i = 0; i < t; i++)
{
scanf("%d", &n);
a[i] = n;
if (t == 1) {
if (i == 0) {
a[i] = (a[i] + 1) * 1;
}
}
else if (t == 2) {
if (i == 0) {
a[i] = (a[i] + 1) * 10;
}
else if (i == 1) {
a[i] = (a[i] + 1) * 1;
}
}
else if (t == 3) {
if (i == 0) {
a[i] = (a[i] + 1) * 100;
}
else if (i == 1) {
a[i] = (a[i] + 1) * 10;
}
else if (i == 2) {
a[i] = (a[i] + 1) * 1;
}
}
else if (t == 4) {
if (i == 0) {
a[i] = (a[i] + 1) * 1000;
}
else if (i == 1) {
a[i] = (a[i] + 1) * 100;
}
else if (i == 2) {
a[i] = (a[i] + 1) * 10;
}
else if (i == 3) {
a[i] = (a[i] + 1) * 1;
}
}
else if (t == 5) {
if (i == 0) {
a[i] = (a[i] + 1) * 10000;
}
else if (i == 1) {
a[i] = (a[i] + 1) * 1000;
}
else if (i == 2) {
a[i] = (a[i] + 1) * 100;
}
else if (i == 3) {
a[i] = (a[i] + 1) * 10;
}
else if (i == 4) {
a[i] = (a[i] + 1) * 1;
}
}
else if (t == 6) {
if (i == 0) {
a[i] = (a[i] + 1) * 100000;
}
else if (i == 1) {
a[i] = (a[i] + 1) * 10000;
}
else if (i == 2) {
a[i] = (a[i] + 1) * 1000;
}
else if (i == 3) {
a[i] = (a[i] + 1) * 100;
}
else if (i == 4) {
a[i] = (a[i] + 1) * 10;
}
else if (i == 4) {
a[i] = (a[i] + 1) * 1;
}
}
}
}
for (int i = 0; i < t; i++)
{
sum = sum + a[i];
}
printf("%d\n", sum);
}

Answer 1

I've reworked on the code from beginning and I've made a solution for you:
#include <stdio.h>
int main(void)
{
int num, sum, remainder, check; // check used as a boolean expression
sum = check = 0;
printf("Enter the sequence: ");
scanf("%d", &num);
while (num > 0)
{
remainder = num % 10; // each time num is reduced
if (remainder != 9)
{
if (check == 0)
sum = (10 * sum) + (remainder + 1);
else
{
sum = (10 * sum) + (remainder + 2);
check = 0;
}
}
else
{
sum = (10 * sum) + 0;
check = 1;
}
num /= 10; // will divide and execute in each iteration until it's true
}
num = sum; // final number will be equal to the sum
sum = 0;
// Summing up the results
while (num > 0)
{
remainder = num % 10;
sum = (10 * sum) + remainder;
num /= 10;
}
printf("Result: %d\n", sum);
return 0;
}
Test Output
Enter the sequence: 23456
Result: 34567
It's just all about the sum & remainder. Hope it helps you understand better.

Answer 2

import java.util.Scanner;
class Main
{
public static void main(String[] args)
{
int num,i=1,j;
Scanner scan=new Scanner(System.in);
int numo=scan.nextInt();num=numo;
for(;numo>0;numo=numo/10,i=i*10)
{
num=num+i;
if(numo%10==9)
num=num-i*10;
}
System.out.println(num);
}
}

Answer 3

The below solution uses the basic remainder and reverse approach:
int addOne(int n)
{
int rem, ans=0, p=1 ;
while(n>0)
{
rem = n%10;
(rem == 9)?rem = 0:rem+=1;
ans+=p*rem;
p*=10;n/=10;
}
return ans;
}
int main() {
int n;
cin>>n;
cout<<addOne(n);
return 0;
}