#include <stdio.h>
int main()
{
int k;
scanf("%d",&k);
for(int i=0;i<k;i++)
{
char num[10];
scanf("%s",&num);
int x = num[0]-'0';
int sum;
int y;
for(int i=0;i<10;i++)
{
if(num[i]=='\0')
{
y = num[i-1]-'0';
sum = x+y;
printf("%d\n",sum);
}
}
}
return 0;
}
This program is to obtain the sum of first and last digit of a given number. I am getting some random values after the output. Can anyone please explain me this and how do I eliminate it?enter image description here
You have an inner loop that looks for the terminating null byte in the array. You find it, but then you keep looping through the array looking for more null bytes. If you happen to find one, it takes whatever the prior byte is and treats it as the last digit.
You can fix this by breaking out of the loop when you find the null terminator:
for(int i=0;i<10;i++)
{
if(num[i]=='\0')
{
y = num[i-1]-'0';
sum = x+y;
printf("%d\n",sum);
break;
}
}
Or by simply using strlen to find the end of the string.
y = num[strlen(num)-1]-'0';
sum = x+y;
printf("%d\n",sum);
The code also doesn't handle reading too many characters, which can overflow the buffer, or reading an empty string, in which case num[strlen(num)-1] is off the start of the array. You can address the former by putting a field width in the scanf format, and the latter with an explicit check:
char num[10];
scanf("%9s",num);
int sum;
if (num[0] == 0) {
sum = 0;
} else {
int x = num[0]-'0';
int y = num[strlen(num)-1]-'0';
sum = x + y;
}
printf("%d\n",sum);
The problem is that you don't stop the inner loop once you print the sum. You continue and use the indeterminate elements in the array num beyond the string terminator. That leads to undefined behavior.
You could solve it either by not having the inner loop (by using index 0 and strlen(num) - 1), or by using break to break out of the loop once you printed the sum.
Related
THE PROBLEM
I am facing the problem when i was solving one of the leetcode question
Find Numbers with Even Number of Digits.
This is the code which i have written, the algorithm that i used here is using a for loop to iterate to all the array elements and then i declared the variables b=0 and l=10.then while using a while loop of condition (c!= nums[i]). The while loops happens until c(which is initialised it to nums[i]) is equal to array element.
then i check whether b%2==0 and increment integer ans
int findNumbers(int* nums, int numsSize){
int ans=0,c;
for (int i = 0; i < numsSize; ++i) {
int b=0,l=10;
while(c!=nums[i])
{
c=nums[i];
c=c%l;
l=l*10;
b++;
}
if(b%2==0 && nums[i]!=49916) {
ans++;
}
}
return ans;
}
If i take two odd digit numbers for suppose [78968,78968]. I am getting the output as 1 while the expected answer should be 0
Output for the above example
Excuse me if you had any difficulties understanding the questions. Learning to write better questions!
#include <stdio.h>
int findNumbers(int* nums, int numsSize){
int ans=0,c;
for (int i = 0; i < numsSize; ++i) {
int b=0,l=10;
c=0;
while(c!=nums[i])
{
c=nums[i];
c=c%l;
l=l*10;
b++;
}
if(b%2==0 && nums[i]!=49916)
{
ans++;
}
}
return ans;
}
int main()
{
int n[]={462,462};
printf("%d",findNumbers(n,2));
return 0;
}
This code should work for you. Actually the problem was when the loop was iterated first time the value of c will be 462 according to above code so while loop will not be iterated as c!=nums[i] will become false and value of b = 0 so if condition will become true so the value of ans will be incremented and you are getting answer as 1.
I need to print values stored in an int array, stopping when a NULL character is encountered ('\0').
So I have this code:
const int display[10] = {1,4,8,2,0,9,2};
int main(){
int i = 0;
for (i = 0; i < 10; i++){
if (display[i] == '\0'){
break;
}
printf("%d\n", display[i]);
}
exit(0);
}
I expected to print all the values of display[10] OR break when a '\0' was encountered but my code breaks on display[4] (0) instead of continuing until display[6].
Any advice on how to achieve this, avoiding printing the null characters at the end of my array?
The null character, '\0', is equal to 0. That's why your loop is only printing the first four elements. It breaks when it encounters 0.
In C, '\0'==0. If you want to print only the initialized fields, put a sentinel (say, a negative number) right after the last initialized field and break the loop when you either encounter the sentinel or count to 10.
const int display[10] = {1,4,8,2,0,9,2,-1 /* a sentinel */};
for (i = 0; i < 10 && display[i] >= 0; i++) {
You do not need to check whether null is exist or not. First declare the array without total number initialization. Then You can just run the loop by checking the total number of array element. Now the question is how you get the number of element? It is easy.
const int arr[]= { 1, 2, 3};
int size = sizeof(arr) / sizeof(arr[0]);
for (int i = 0; i < size; i++) {
printf("%d", arr[i]);
}
I'm trying to generate a random 10-digit code, but even though I use the absolute value of every number in the code, it still sometimes prints a negative value
#include <stdio.h>
int main()
{
int i;
int r;
int barcode[11];
srand(time(NULL));
for(i=0;i <= 10;i++){
r = rand() % 10;
barcode[i] = abs(r);
}
printf("%d",barcode);
return 0;
}
Because you are actually printing the address of an integer array, not a string.
This line:
printf("%d",barcode);
Basically prints the address of barcode as a signed integer instead of the contents of barcode.
You of course could do this:
printf("%d%d%d%d%d%d%d%d%d%d",barcode[0], barcode[1], barcode[2], barcode[3], barcode[4], barcode[5], barcode[6], barcode[7], barcode[8], barcode[9]);
But perhaps a better way is to generate a string of characters instead of an array of integers. Quick mod to your code is to add to '0' to each random value in each interation of the loop and append to a char array.
int main()
{
int i;
int r;
char barcode[11]; // array of chars instead of ints
srand(time(NULL));
for(i=0; i < 10; i++) // loop 10 times, not 11
{
r = rand() % 10;
barcode[i] = '0' + r; // convert the value of r to a printable char
}
barcode[10] = '\0'; // null terminate your string
printf("%s\n",barcode);
return 0;
}
The above will generate a 10 digit code, with a small possibility of the first number being a leading zero. If that's not what you want, that's a simple bug fix. (Which I'll leave up to you...)
i got a problem which i can't solve
I want to know all prime numbers below a given limit x. Allowing me to enter x and calculate the prime numbers using the method of Erastosthenes. Displaying the result on the screen and saving it to a text file.
Calculating the primenumbers below the x, printing them and saving them to a text file worked, the only problem i have is that x can't exceed 500000
could you guys help me?
#include <stdio.h>
#include <math.h>
void sieve(long x, int primes[]);
main()
{
long i;
long x=500000;
int v[x];
printf("give a x\n");
scanf("%d",&x);
FILE *fp;
fp = fopen("primes.txt", "w");
sieve(x, v);
for (i=0;i<x;i++)
{
if (v[i] == 1)
{
printf("\n%d",i);
fprintf(fp, "%d\n",i);
}
}
fclose(fp);
}
void sieve(long x, int primes[])
{
int i;
int j;
for (i=0;i<x;i++)
{
primes[i]=1; // we initialize the sieve list to all 1's (True)
primes[0]=0,primes[1]=0; // Set the first two numbers (0 and 1) to 0 (False)
}
for (i=2;i<sqrt(x);i++) // loop through all the numbers up to the sqrt(n)
{
for (j=i*i;j<x;j+=i) // mark off each factor of i by setting it to 0 (False)
{
primes[j] = 0;
}
}
}
You will be able to handle four times as many values by declaring char v [500000] instead of int v [100000].
You can handle eight times more values by declaring unsigned char v [500000] and using only a single bit for each prime number. This makes the code a bit more complicated.
You can handle twice as many values by having a sieve for odd numbers only. Since 2 is the only even prime number, there is no point keeping them in the sieve.
Since memory for local variables in a function is often quite limited, you can handle many more values by using a static array.
Allocating v as an array of int is wasteful, and making it a local array is risky, stack space being limited. If the array becomes large enough to exceed available stack space, the program will invoke undefined behaviour and likely crash.
While there are ways to improve the efficiency of the sieve by changing the sieve array to an array of bits containing only odd numbers or fewer numbers (6n-1 and 6n+1 is a good trick), you can still improve the efficiency of your simplistic approach by a factor of 10 with easy changes:
fix primes[0] and primes[1] outside the loop,
clear even offsets of prime except the first and only scan odd numbers,
use integer arithmetic for the outer loop limit,
ignore numbers that are already known to be composite,
only check off odd multiples of i.
Here is an improved version:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void sieve(long x, unsigned char primes[]) {
long i, j;
for (i = 0; i < x; i++) {
primes[i] = i & 1;
}
primes[1] = 0;
primes[2] = 1;
/* loop through all odd numbers up to the sqrt(x) */
for (i = 3; (j = i * i) < x; i += 2) {
/* skip composite numbers */
if (primes[i] == 0)
continue;
/* mark each odd multiple of i as composite */
for (; j < x; j += i + i) {
primes[j] = 0;
}
}
}
int main(int argc, char *argv[]) {
long i, x, count;
int do_count = 0;
unsigned char *v;
if (argc > 1) {
x = strtol(argv[1], NULL, 0);
} else {
printf("enter x: ");
if (scanf("%ld", &x) != 1)
return 1;
}
if (x < 0) {
x = -x;
do_count = 1;
}
v = malloc(x);
if (v == NULL) {
printf("Not enough memory\n");
return 1;
}
sieve(x, v);
if (do_count) {
for (count = i = 0; i < x; i++) {
count += v[i];
}
printf("%ld\n", count);
} else {
for (i = 0; i < x; i++) {
if (v[i] == 1) {
printf("%ld\n", i);
}
}
}
free(v);
return 0;
}
I believe the problem you are having is allocating an array of int if more than 500000 elements on the stack. This is not an efficient way, to use an array where the element is the number and the value indicates whether it is prime or not. If you want to do this, at least use bool, not int as this should only be 1 byte, not 4.
Also notice this
for (i=0;i<x;i++)
{
primes[i]=1; // we initialize the sieve list to all 1's (True)
primes[0]=0,primes[1]=0; // Set the first two numbers (0 and 1) to 0 (False)
}
You are reassigning the first two elements in each loop. Take it out of the loop.
You are initializing x to be 500000, then creating an array with x elements, thus it will have 500000 elements. You are then reading in x. The array will not change size when the value of x changes - it is fixed at 500000 elements, the value of x when you created the array. You want something like this:
long x=500000;
printf("give a x\n");
scanf("%d",&x);
int *v = new int[x];
This fixes your fixed size array issue, and also gets it off the stack and into the heap which will allow you to allocate more space. It should work up to the limit of the memory you have available.
So on Project Euler the Problem 4 states the following:
A palindromic number reads the same
both ways. The largest palindrome made
from the product of two 2-digit
numbers is 9009 = 91 99.
Find the largest palindrome made from
the product of two 3-digit numbers.
I have tried the following:
#include <stdio.h>
#include <stdlib.h>
int check(int result)
{
char b[7];
sprintf(b, "%d", result);
if (b[0] == b[5] && b[1] == b[4] && b[2] == b[3])
{
return 1;
}
else
{
return 0;
}
}
int main () {
int i;
int g;
int final;
for (i = 999; i > 99; i--)
{
for (g = 999; g > 99; g--)
{
if (check(g*i) == 1)
{
final = g*i;
goto here;
}
}
}
here:
printf("%d", final);
}
But, this does not work. Instead of the right answer, I get 580085, which I guess is a palindrome at least, but still not the right answer.
Let me explain my program starting from int main:
int i and int g are my multipliers. They are those two three digit numbers.
int final is the number that will store the largest palindrome.
I start two for loops going to down to get every number possibility.
I get out of the loop using a goto when the first palindrome is reached(probably should not but, it doesn't effect a small program like this too much).
The first palindrome should be the biggest one possible since I am counting down from the top.
Let me now explain my check:
First off since these are two three digit numbers multiplying together to determine the size a char would need to be to hold that value I went to a calculator and multiplied 999 * 999 and it ended up being 6 then I need to add one because I found out from one the questions I posted earlier that sprintf puts a \0 character at the end.
Ok, now that I have a char and all, I copied result (which i*g in int main) and put it in char b[7].
Then I just checked b to see if it equalled it self with by hard coding each slot I needed to check for.
Then I returned accordingly, 1 for true, and 2 for false.
This seems perfectly logical to me but, it does not work for some weird reason. Any hints?
This assumption is wrong:
The first palindrome should be the biggest one possible since I am counting down from the top.
You will check 999*100 = 99900 before 998*101 = 100798, so clearly you can´t count on that.
The problem is that the first palindrome that you find is not the bigger one for sure.
Just an example:
i = 900, g = 850 -> 765000
i = 880, g = 960 -> 844800
The first one is smaller, but since you iterate first on i, then on g it will be discovered first.
Ok, they are not palindrome but the concept is the same..
I think you are tackling this problem back to front. It would be more efficient to generate the palindromes from highest to lowest then check by factorizing them. First one that has two three digit factors is the answer.
e.g.
bool found = false;
for (int i = 998; i >= 100; i--)
{
char j[7];
sprintf(j,"%d",i);
j[3]= j[2];
j[4]= j[1];
j[5]= j[0];
int x =atoi(j);
int limit = sqrt((float) x);
for (int z = 999; z >= limit; z--)
{
if (x%z==0){
printf("%d",x);
found = true;
break;
}
}
if (found) break;
}
The first palindrome should be the biggest one possible since I am counting down from the top
The problem is that you might have found a palindrome for a large i and a small g. It's possible that there's a larger palindrome that's the product of j and k where:
i > j and
g < k
(I hope this makes sense).
Java Implementation:
public class Palindrome {
public static void main(String[] args)
{ int i, j;
int m = 1;
int k =11;
boolean flag = false;
while (true)
{;
if (flag) j = m + 1;
else j = m;
for (i = k; i > 0; i--)
{
j++;
int number, temp, remainder, sum = 0;
number = temp = (1000 - i) * (1000 - j);
while (number > 0)
{
remainder = number % 10;
number /= 10;
sum = sum * 10 + remainder;
}
if (sum == temp)
{
System.out.println("Max value:"+temp);
return;
}
}
if (flag)
m++;
k=k+11;
flag = !flag;
}
}
}
A word on performance. You have the possibility of duplicating many of the products because you are using a pretty simple nested loop approach. For instance, you start with 999*999 and then 999*998, etc. When the inner loop finishes, you will decrement the outer loop and start again with 998*999, which is the same as 999*998.
Really, what you want to do is start the inner loop with the same value as the current outer loop value. This will eliminate your duplicate operations. Something like this...
for (i = 999; i > 99; i--)
{
for (g = i; g > 99; g--)
{
...
However, as Emilio pointed out, your assumption that the first palindrome you find will be the answer is incorrect. You need to compute the biggest numbers first, obviously. So you should try them in this order; 999*999, 999*998, 998*998, 999*997, 998*997, etc...
Haven't tested it but I think you want something like this (pseudo code):
x = 999;
n = 0;
while (++n <= x)
{
j = x;
k = j - n;
while (j >= k)
{
y = j-- * k;
if (check(y))
stop looking
}
}
I found this article which might help you. It has improved brute force approach.
All the above provided answers are excellent, but still I could not restrict myself from writing the code. The code posted by #thyrgle is absolutely perfect. Only a slight correction which he needs to do is just check which product is the maximum.
The code can be as
int i,j,max=0,temp;
for(i=999;i>=100;i--){
for(j=i;j>=100;j--){
temp=i*j;
if(isPalin(temp) && temp>max){
max=temp;
}
}
}
cout<<max<<"\n";
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int a[6];
void convertToString(int xy){
int i,t=100000;
for(i=0;i<6;i++){
a[i]=xy/t;
xy = xy % t;
t=t/10;
}
}
int check(){
int i;
for(i=0;i<3;i++){
if(a[i]!=a[6-i]){
return 0;
}
}
return 1;
}
void main(){
int x,y,xy,status=0;
int i=0,j=0,p=0;
for(x=999;x>99;x--){
for(y=x;y>99;y--){
xy=x*y;
convertToString(xy);
status = check();
if(status==1){
if(xy>p){
p=xy;
i=x;
j=y;
}
}
}
}
printf("\nTwo numbers are %d & %d and their product is %d",i,j,p);
}
x,y=999,999
k=0
pal=[]
while (y>99):
while (x>=100):
m=x*y
n=x*y
while (n!=0):
k=k*10+(n%10)
n=int(n/10)
if(m==k):
if k not in pal:
pal.append(k)
x=x-1
k=0
else:
y,x=y-1,999
pal.sort()
print(pal)
it gives 906609 as the largest palindrome number