I was doing a programming question and one of the sample output is 64197148392731290. My code for that question is correct as it is giving me the right answers for other test cases (output for those test cases are in single digit).
I understand that there will be too many iterations for the test case which has 64197148392731290 as output. So what should I do to get correct answer for that test case too.
Here is the code :
#include<stdio.h>
#include<string.h>
int main() {
int test_case;long long int i, j, count, n, k, k1;
scanf("%d", &test_case);
while(test_case--) {
scanf("%lld%lld", &n, &k);
char a[n];
count=0;
k1=k;
scanf("%s", a);
while(k1--) {
strcat(a,a);
}
for(i=0;i<(n*k);i++) {
if(a[i]=='a') {
for(j=(i+1);j<(n*k);j++) {
if(a[j]=='b') {
count++;
}
}
}
}
printf("%lld\n", count);
}
return 0;
}
Sample Input and Output :
Input:
3
4 2
abcb
7 1
aayzbaa
12 80123123
abzbabzbazab
Output:
6
2
64197148392731290
My task is to count the number of subsequences "ab" (not necessarily consecutive) in the new string. The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
The first line of each test case contains two integers N and K, denoting the length of the initial string S and the number of repetitions respectively.
The second line contains a string S. Its length is exactly N, and each of its characters is a lowercase English letter.
If you are trying to store input in "int" that wont work coz this number its out of range, change it to "long long int"
Well the previous answer was sure wrong. Thanks for the code.
Sorry don't have time for detailed study but preliminary analysis tells me that maybe the error is because you are trying to store a sting of length 2n in a[n]. It works for smaller values since when you declare
char a[n];
^
variable known at runtime
it actually allocates a large block so that any value of n within range is possible. For large values strcat(a,a) will probably fail.
Basically somewhere down the line the string becomes corrupt. Most probably that is because of strcat. I suggest remove strcat, do something else to a similar effect.
Related
I'm trying to create a complete C program to read ten alphabets and display them on the screen. I shall also have to find the number of a certain element and print it on the screen.
#include <stdio.h>
#include <conio.h>
void listAlpha( char ch)
{
printf(" %c", ch);
}
int readAlpha(){
char arr[10];
int count = 1, iterator = 0;
for(int iterator=0; iterator<10; iterator++){
printf("\nAlphabet %d:", count);
scanf(" %c", &arr[iterator]);
count++;
}
printf("-----------------------------------------");
printf("List of alphabets: ");
for (int x=0; x<10; x++)
{
/* I’m passing each element one by one using subscript*/
listAlpha(arr[x]);
}
printf("%c",arr);
return 0;
}
int findTotal(){
}
int main(){
readAlpha();
}
The code should be added in the findTotal() element. The output is expected as below.
Output:
List of alphabets : C C C A B C B A C C //I've worked out this part.
Total alphabet A: 2
Total alphabet B: 2
Total alphabet C: 6
Alphabet with highest hit is C
I use an array to count the number of the existence of each character,
I did this code but the display of number of each character is repeated in the loop
int main()
{
char arr[100];
printf("Give a text :");
gets(arr);
int k=strlen(arr);
for(int iterator=0; iterator<k; iterator++)
{
printf("[%c]",arr[iterator]);
}
int T[k];
for(int i=0;i<k;i++)
{
T[i]=arr[i];
}
int cpt1=0;
char d;
for(int i=0;i<k;i++)
{int cpt=0;
for(int j=0;j<k;j++)
{
if(T[i]==T[j])
{
cpt++;
}
}
if(cpt>cpt1)
{
cpt1=cpt;
d=T[i];
}
printf("\nTotal alphabet %c : %d \n",T[i],cpt);
}
printf("\nAlphabet with highest hit is : %c\n",d,cpt1);
}
There is no way to get the number of elements You write in an array.
Array in C is just a space in the memory.
C does not know what elements are actual data.
But there are common ways to solve this problem in C:
as mentioned above, create an array with one extra element and, fill the element after the last actual element with zero ('\0'). Zero means the end of the actual data. It is right if you do not wish to use '\0' among characters to be processed. It is similar to null-terminated strings in C.
add the variable to store the number of elements in an array. It is similar to Pascal-strings.
#include <stdio.h>
#include <string.h>
#define ARRAY_SIZE 10
char array[ARRAY_SIZE + 1];
int array_len(char * inp_arr) {
int ret_val = 0;
while (inp_arr[ret_val] != '\0')
++ret_val;
return ret_val;
}
float array_with_level[ARRAY_SIZE];
int array_with_level_level;
int main() {
array[0] = '\0';
memcpy(array, "hello!\0", 7); // 7'th element is 0
printf("array with 0 at the end\n");
printf("%s, length is %d\n", array, array_len(array));
array_with_level_level = 0;
const int fill_level = 5;
int iter;
for (iter = 0; iter < fill_level; ++iter) {
array_with_level[iter] = iter*iter/2.0;
}
array_with_level_level = iter;
printf("array with length in the dedicated variable\n");
for (int i1 = 0; i1 < array_with_level_level; ++i1)
printf("%02d:%02.2f ", i1, array_with_level[i1]);
printf(", length is %d", array_with_level_level);
return 0;
}
<conio.h> is a non-standard header. I assume you're using Turbo C/C++ because it's part of your course. Turbo C/C++ is a terrible implementation (in 2020) and the only known reason to use it is because your lecturer made you!
However everything you actually use here is standard. I believe you can remove it.
printf("%c",arr); doesn't make sense. arr will be passed as a pointer (to the first character in the array) but %c expects a character value. I'm not sure what you want that line to do but it doesn't look useful - you've listed the array in the for-loop.
I suggest you remove it. If you do don't worry about a \0. You only need that if you want to treat arr as a string but in the code you're handling it quite validly as an array of 10 characters without calling any functions that expect a string. That's when it needs to contain a 0 terminator.
Also add return 0; to the end of main(). It means 'execution successful' and is required to be conformant.
With those 3 changes an input of ABCDEFGHIJ produces:
Alphabet 1:
Alphabet 2:
Alphabet 3:
Alphabet 4:
Alphabet 5:
Alphabet 6:
Alphabet 7:
Alphabet 8:
Alphabet 9:
Alphabet 10:-----------------------------------------List of alphabets: A B C D E F G H I J
It's not pretty but that's what you asked for and it at least shows you've successfully read in the letters. You may want to tidy it up...
Remove printf("\nAlphabet %d:", count); and insert printf("\nAlphabet %d: %c", count,arr[iterator]); after scanf(" %c", &arr[iterator]);.
Put a newline before and after the line of minus signs (printf("\n-----------------------------------------\n"); and it looks better to me.
But that's just cosmetics. It's up to you.
There's a number of ways to find the most frequent character. But at this level I recommend a simple nested loop.
Here's a function that finds the most common character (rather than the count of the most common character) and if there's a tie (two characters with the same count) it returns the one that appears first.
char findCommonest(const char* arr){
char commonest='#'; //Arbitrary Bad value!
int high_count=0;
for(int ch=0;ch<10;++ch){
const char counting=arr[ch];
int count=0;
for(int c=0;c<10;++c){
if(arr[c]==counting){
++count;
}
}
if(count>high_count){
high_count=count;
commonest=counting;
}
}
return commonest;
}
It's not very efficient and you might like to put some printfs in to see why!
But I think it's at your level of expertise to understand. Eventually.
Here's a version that unit-tests that function. Never write code without a unit test battery of some kind. It might look like chore but it'll help debug your code.
https://ideone.com/DVy7Cn
Footnote: I've made minimal changes to your code. There's comments with some good advice that you shouldn't hardcode the array size as 10 and certainly not litter the code with that value (e.g. #define ALPHABET_LIST_SIZE (10) at the top).
I have used const but that may be something you haven't yet met. If you don't understand it and don't want to learn it, remove it.
The terms of your course will forbid plagiarism. You may not cut and paste my code into yours. You are obliged to understand the ideas and implement it yourself. My code is very inefficient. You might want to do something about that!
The only run-time problem I see in your code is this statement:
printf("%c",arr);
Is wrong. At this point in your program, arr is an array of char, not a single char as expected by the format specifier %c. For this to work, the printf() needs to be expanded to:
printf("%c%c%c%c%c%c%c%c%c%c\n",
arr[0],arr[1],arr[2],arr[3],arr[4],
arr[5],arr[6],arr[7],arr[8],arr[9]);
Or: treat arr as a string rather than just a char array. Declare arr as `char arr[11] = {0};//extra space for null termination
printf("%s\n", arr);//to print the string
Regarding this part of your stated objective:
"I shall also have to find the number of a certain element and print it on the screen. I'm new to this. Please help me out."
The steps below are offered to modify the following work
int findTotal(){
}
Change prototype to:
int FindTotal(char *arr);
count each occurrence of unique element in array (How to reference)
Adapt above reference to use printf and formatting to match your stated output. (How to reference)
Here is the code for "The Next Palindrome" which I wrote in C:
#include<stdio.h>
int main(void)
{
int check(int); //function declaration
int t,i,k[1000],flag,n;
scanf("%d",&t); //test cases
for(i=0; i<t; i++)
scanf("%d",&k[i]); //numbers
for(i=0; i<t; i++)
{
if(k[i]<=9999999) //Number should be of 1000000 digits
{
k[i]++;
while(1)
{
flag=check(k[i]); //palindrome check
if(flag==1)
{
printf("%d\n",k[i]); //prints if it is palindrome and breaks
break;
}
else
k[i]++; //go to the next number
}
}
}
return 0;
}
int check(int n)
{
int rn=0;
int temp=n;
while(n!=0)
{
rn=rn*10+n%10; //reversing
n=n/10;
}
if(rn==temp) //number is palindrome
return 1;
else //number is not a palindrome
return 0;
}
It is a beginner level problem from SPOJ.
I tried to run this code on Codeblocks and it ran fluently.
In SPOJ, why is it showing wrong output?
In SPOJ, why is it showing wrong output?
This is nice solution and it works for small inputs, however it will not pass SPOJ for several reasons.
The requirement is:
A positive integer is called a palindrome if its representation in the
decimal system is the same when read from left to right and from right
to left. For a given positive integer K of not more than 1000000
digits, write the value of the smallest palindrome larger than K to
output. Numbers are always displayed without leading zeros.
Input:
The first line contains integer t, the number of test cases.
Integers K are given in the next t lines.
So which requirements are broken in your program?
1) Your assumption is that only 1000 numbers will be given for processing since
you declared
k[1000]
wrong, the number of lines is given in first line. It could be much more than 1000. You have to dynamically assign the storage for the numbers.
2)
The line
if(k[i]<=9999999)
assumes that input is less than 9999999
- wrong, the requirement says positive integer K of not more than 1000000 digits which imply that much larger numbers e.g. 199999991 also have to be accepted.
3) The statement
For a given positive integer K of not more than 1000000 digits
as well as warning
Warning: large Input/Output data, be careful with certain languages
leads us to conclusion that really big numbers should be expected!
The int type is not a proper vehicle for storing such big numbers. The int will fail to hold the value if the number is bigger than INT_MAX +2147483647. (Check C Library <limits.h>)
So, how to pass SPOJ challange?
Hint:
One of the possible solutions - operate on strings.
I am following the following function to calculate factorials of big numbers link, and I would like to understand a bit more why some things are happening...
#include<stdio.h>
#define MAX 10000
void factorialof(int);
void multiply(int);
int length = 0;
int fact[MAX];
int main(){
int num;
int i;
printf("Enter any integer number : ");
scanf("%d",&num);
fact[0]=1;
factorialof(num);
printf("Factorial is : ");
for(i=length;i>=0;i--){
printf("%d",fact[i]);
}
return 0;
}
void factorialof(int num){
int i;
for(i=2;i<=num;i++){
multiply(i);
}
}
void multiply(int num){
long i,r=0;
int arr[MAX];
for(i=0;i<=length;i++){
arr[i]=fact[i];
}
for(i=0;i<=length;i++){
fact[i] = (arr[i]*num + r)%10;
r = (arr[i]*num + r)/10;
//printf("%d ",r);
}
if(r!=0){
while(r!=0){
fact[i]=r%10;
r= r/10;
i++;
}
}
length = i-1;
}
My questions are:
What is the real meaning of the MAX constant? What does it mean if it's bigger or smaller?
I have found out that if I have a MAX = 10000 (as in the example), I can calculate up to 3250! If I try with 3251! I get a 'Abort trap: 6' message. Why is that number? Where does it come from?
Which would be the difference if I compile this code for a 32-bit machine with the flag -m32? Would it run he same as in 64-bit?
Thanks!
As Scott Hunter points out, MAX is the maximum number of elements in the fact and arr arrays, which means it's the maximum number of digits that can occur in the result before the program runs out of space.
Note that the code only uses MAX in its array declarations. Nowhere does it use MAX to determine whether or not it's trying to read from or write to memory beyond the end of those arrays. This is a Bad Thing™. Your "Abort trap: 6" error is almost certainly occurring because trying to compute 3251! is doing exactly that: using a too-large index with arr and fact.
To see the number of digits required for a given factorial, you can increase MAX (say, to 20,000) and replace the existing printf calls in main with something like this:
printf("Factorial requires %d digits.\n", length + 1);
Note that I use length + 1 because length isn't the number of digits by itself: rather, it's the index of the array position in fact that contains the most-significant digit of the result. If I try to compute 3251!, the output is:
Factorial requires 10008 digits.
This is eight digits more than you have available in fact with the default MAX value of 10,000. Once the program logic goes beyond the allocated space in the array, its behavior is undefined. You happen to be seeing the error "Abort trap: 6."
Interestingly, here's the output when I try to compute 3250!:
Factorial requires 10005 digits.
That's still too many for the program to behave reliably when MAX is set to 10,000, so the fact that your program calculates 3250! successfully might be surprising, but that's the nature of undefined behavior: maybe your program will produce the correct result, maybe it will crash, maybe it will become self-aware and launch its missiles against the targets in Russia (because it knows that the Russian counterattack will eliminate its enemies over here). Coding like this is not a good idea. If your program requires more space than it has available in order to complete the calculation, it should stop and display an appropriate error message rather than trying to continue what it's doing.
MAX is the number of elements in fact and arr; trying to access an element with an index >= MAX would be bad.
Error messages are often specific to the environment you are using, which you have provided no details for.
They are not the same, but the differences (for example, the size of pointers) should not affect this code in any discernible way.
I have just started competitive programming in SPOJ.I'm confused from sometime why i'm getting runtime error in ideone.The question is:
A positive integer is called a palindrome if its representation in the decimal system is the same when read from left to right and from right to left. For a given positive integer K of not more than 1000000 digits, write the value of the smallest palindrome larger than K to output. Numbers are always displayed without leading zeros.
Input
The first line contains integer t, the number of test cases. Integers K are given in the next t lines.
Output
For each K, output the smallest palindrome larger than K.
Example
Input:
2
808
2133
Output:
818
2222
My program:
#include <stdio.h>
int main(void)
{
int t,i,reverse,same;
scanf("%d",&t); //t is no. of test cases
int num[t]; //num[t] is array of t elements
for(i=0;i<t;i++)
scanf("%d",&num[i]);
i=0; //since i will be equal to t therefore i is assigned to 0.
while(t--)
{
if(num[i]<=1000000)
{
while(num[i]++)
{
reverse=0;
same=num[i];
while(same>0)
{
reverse=reverse*10;
reverse=reverse+same%10;
same=same/10;
}
if(reverse==num[i])
printf("%d",reverse);
printf("\n");
if(reverse==num[i])
break;
}
}
i++;
}
return 0;
}
I don't know where i'm wrong.I'm sorry i'm asking this question may this question is asked by someone before.I tried to find the result but could not get the answer.Thankyou in advance and sorry for my bad english.
The question doesn't say that the number will be less than 1000000. It says that the number has less than 1 million digits. A number with a million digits looks like this
591875018734106743196734198673419067843196874398674319687431986743918674319867431986743198674319876341987643198764319876341987643198764319876431987643198763419876431987643198764319876139876...
You can't use scanf to read a number that has a million digits, and you can't store that number in an int.
The most likely reason for your error to occur is some memory fault. Keep in mind that online judges/compilers limit your available memory and if you try to allocate/use more memory than available, you get a runtime error. This also happens on your machine, but usually you have a lot more memory available for your program than in the case of online judges.
In your case, you could reduce the memory usage of your program by changing the data type of the num array from int to something like short or even char.
This is a homework project I was assigned some time ago... I've been successful in getting this far on my own, and the only hiccup I have left is (I believe) an issue with data types and overflow.
I've tried changing over to unsigned and double, and the code complies and still accepts input in the terminal, but it seems to hang up after that... nothing is printed and it looks like it's caught in a loop.
Here is the code...
/* pascaltri.c
* A program that takes a single integer as input and returns the nth line of
* Pascal's Triangle. Uses factorial() function to help find items of
* individual entries on a given row.
*/
#include <stdio.h>
#include <stdlib.h>
long factorial(long i)
{
long fact = 1;
while(i > 1)
{
fact = fact * i;
i = i - 1;
}
return fact;
}
main(void)
{
long n;
long *nPtr;
nPtr = &n;
scanf(" %i", nPtr);
if (n >= 0)
{
long k;
long *kPtr;
kPtr = &k;
for(k = 0; k <= n; k++)
{
long ans;
long *ansPtr;
ansPtr = &ans;
ans = factorial(n) / (factorial(k) * factorial(n - k));
printf("\n %i", ans);
}
return 0;
}
return 0;
}
It's not perfect or pretty, but it works up to an input of 13 (that is, row 14) of the triangle. Beyond that I start getting gibberish and even negative values sprinkled throughout the returns... much larger values break the code and return nothing but an exit error message.
Any ideas on how I can correct this problem? I've been staring at the screen for much to long to really see anything myself. Also, it's not essential, but I would like to print my return values on one line, rather than having them separated by a newline character.
1 5 10 10 5 1
Would the easiest way be to load the values into an array as they are computed, and then print the array? Or is there a built-in way I can tell the print statement to occur on only one line?
You are suffering from integer overflow. You may need to find a different approach to the algorithm to avoid having to calculate the large numbers.
In answer to your other point about the newline, you are explicitly printing the newline with the \n in your print statement. Remove it, and you will get answers printed on one line. You probably want to inlucde a final printf("\n"); at the end so the whole line is terminated in a newline.
Some other observations:
You don't need the first return 0; - the control will drop out of
the bottom of the if block and on to the second (should be only)
return 0; and not cause any problems.
You're declaring kPtr but not using it anywhere
You don't need to declare a separate variable nPtr to pass to scanf; you can pass &n directly.
For the garbage, you are most likely running into an integer overflow, that is, your calculated values become too large for the long data type. You should correct it by calculating your factorial function without explicitely calculating n!.
Change scanf(" %i", nPtr); to
scanf(" %ld", nPtr);
and printf("\n %i", ans); to
printf("\n %ld", ans);
to get printout on one line, use:
printf(" %ld", ans);
If you are using gcc, turn on warnings, i.e. use -Wall.