I have a code which reads around (10^5) int(s) from stdin and then after performing ## i output them on stdout. I have taken care of the INPUT part by using "setvbuf" & reading lines using "fgets_unlocked()" and then parsing them to get the required int(s).
I have 2 issues which i am not able to come over with:
1.) As i am printing int(s) 5 million on stdout its taking lot of time : IS THERE ANY WAY TO REDUCE THIS( i tried using fwrite() but the o/p prints unprintable characters due to the reason using fread to read into int buffer)
2.) After parsing the input for the int(s) say 'x' i actually find the no of divisors by doing %(mod) for the no in a loop.(See in the code below): Maybe this is also a reason for my code being times out:
Any suggestions on this to improved.
Many thanks
This is actually a problem from http://www.codechef.com/problems/PD13
# include <stdio.h>
# define SIZE 32*1024
char buf[SIZE];
main(void)
{
int i=0,chk =0;
unsigned int j =0 ,div =0;
int a =0,num =0;
char ch;
setvbuf(stdin,(char*)NULL,_IOFBF,0);
scanf("%d",&chk);
while(getchar_unlocked() != '\n');
while((a = fread_unlocked(buf,1,SIZE,stdin)) >0)
{
for(i=0;i<a;i++)
{
if(buf[i] != '\n')
{
num = (buf[i] - '0')+(10*num);
}
else
if(buf[i] == '\n')
{
div = 1;
for(j=2;j<=(num/2);j++)
{
if((num%j) == 0) // Prob 2
{
div +=j;
}
}
num = 0;
printf("%d\n",div); // problem 1
}
}
}
return 0;
}
You can print far faster than printf.
Look into itoa(), or write your own simple function that converts integers to ascii very quickly.
Here's a quick-n-dirty version of itoa that should work fast for your purposes:
char* custom_itoa(int i)
{
static char output[24]; // 64-bit MAX_INT is 20 digits
char* p = &output[23];
for(*p--=0;i/=10;*p--=i%10+0x30);
return ++p;
}
note that this function has some serious built in limits, including:
it doesn't handle negative numbers
it doesn't currently handle numbers greater than 23-characters in decimal form.
it is inherently thread-dangerous. Do not attempt in a multi-threaded environment.
the return value will be corrupted as soon as the function is called again.
I wrote this purely for speed, not for safety or convenience.
Version 2 based on suggestion by #UmNyobe and #wildplasser(see above comments)
The code execution took 0.12 seconds and 3.2 MB of memory on the online judge.
I myself checked with 2*10^5 int(input) in the range from 1 to 5*10^5 and the execution took:
real 0m0.443s
user 0m0.408s
sys 0m0.024s
**Please see if some more optimization can be done.
enter code here
/** Solution for the sum of the proper divisor problem from codechef **/
/** # author dZONE **/
# include <stdio.h>
# include <math.h>
# include <stdlib.h>
# include <error.h>
# define SIZE 200000
inline int readnum(void);
void count(int num);
int pft[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709};
unsigned long long int sum[SIZE];
int k = 0;
inline int readnum(void)
{
int num = 0;
char ch;
while((ch = getchar_unlocked()) != '\n')
{
if(ch >=48 && ch <=57)
{
num = ch -'0' + 10*num;
}
}
if(num ==0)
{
return -1;
}
return num;
}
void count(int num)
{
unsigned int i = 0;
unsigned long long tmp =0,pfac =1;
int flag = 0;
tmp = num;
sum[k] = 1;
for(i=0;i<127;i++)
{
if((tmp % pft[i]) == 0)
{
flag =1; // For Prime numbers not in pft table
pfac =1;
while(tmp % pft[i] == 0)
{
tmp =tmp /pft[i];
pfac *= pft[i];
}
pfac *= pft[i];
sum[k] *= (pfac-1)/(pft[i]-1);
}
}
if(flag ==0)
{
sum[k] = 1;
++k;
return;
}
if(tmp != 1) // For numbers with some prime factors in the pft table+some prime > 705
{
sum[k] *=((tmp*tmp) -1)/(tmp -1);
}
sum[k] -=num;
++k;
return;
}
int main(void)
{
int i=0,terms =0,num = 0;
setvbuf(stdin,(char*)NULL,_IOFBF,0);
scanf("%d",&terms);
while(getchar_unlocked() != '\n');
while(terms--)
{
num = readnum();
if(num ==1)
{
continue;
}
if(num == -1)
{
perror("\n ERROR\n");
return 0;
}
count(num);
}
i =0;
while(i<k)
{
printf("%lld\n",sum[i]);
++i;
}
return 0;
}
//Prob 2 Is your biggesr issue right now.... You just want to find the number of divisors?
My first suggestion will be to cache your result to some degree... but this requires potentially twice the amount of storage you have at the beginning :/.
What you can do is generate a list of prime numbers before hand (using the sieve algorithm). It will be ideal to know the biggest number N in your list and generate all primes till his square root. Now for each number in your list, you want to find his representation as product of factors, ie
n = a1^p1 * a1^p2 *... *an^pn
Then the sum of divisors will be.
((a1^(p1+1) - 1)/(a1 - 1))*((a2^(p2+1) - 1)/(a2-1))*...*((an^(pn+1) - 1)/(an-1))
To understand you have (for n = 8) 1+ 2 + 4 + 8 = 15 = (16 - 1)/(2 - 1)
It will drastically improve the speed but integer factorization (what you are really doing) is really costly...
Edit:
In your link the maximum is 5000000 so you have at most 700 primes
Simple decomposition algorithm
void primedecomp(int number, const int* primetable, int* primecount,
int pos,int tablelen){
while(pos < tablelen && number % primetable[pos] !=0 )
pos++;
if(pos == tablelen)
return
while(number % primetable[pos] ==0 ){
number = number / primetable[pos];
primecount[pos]++;
}
//number has been modified
//too lazy to write a loop, so recursive call
primedecomp(number,primetable,primecount, pos+1,tablelen);
}
EDIT : rather than counting, compute a^(n+1) using primepow = a; primepow = a*primepow;
It will be much cleaner in C++ or java where you have hashmap. At the end
primecount contains the pi values I was talking about above.
Even if it looks scary, you will create the primetable only once. Now this algorithm
run in worst case in O(tablelen) which is O(square root(Nmax)). your initial
loop ran in O(Nmax).
Related
this is my code, I want to make a function that when it is called will generate a number between 1111 to 9999, I don't know how to continue or if I've written this right. Could someone please help me figure this function out. It suppose to be simple.
I had to edit the question in order to clarify some things. This function is needed to get 4 random digits that is understandable from the code. And the other part is that i have to make another function which is a bool. The bool needs to first of get the numbers from the function get_random_4digits and check if there contains a 0 in the number. If that is the case then the other function, lets call it unique_4digit, should disregard of that number that contained a 0 in it and check for a new one to use. I need not help with the function get_random_4digitsbecause it is correct. I need helt constructing a bool that takes get_random_4digits as an argument to check if it contains a 0. My brain can't comprehend how I first do the get_random_4digit then pass the answer to unique_4digits in order to check if the random 4 digits contains a 0 and only make it print the results that doesn't contain a 0.
So I need help with understanding how to check the random 4 digits for the integer 0 and not let it print if it has a 0, and only let the 4 random numbers print when it does not contain a 0.
the code is not suppose to get more complicated than this.
int get_random_4digit(){
int lower = 1000, upper = 9999,answer;
answer = (rand()%(upper-lower)1)+lower;
return answer;
}
bool unique_4digits(answer){
if(answer == 0)
return true;
if(answer < 0)
answer = -answer;
while(answer > 0) {
if(answer % 10 == 0)
return true;
answer /= 10;
}
return false;
}
printf("Random answer %d\n", get_random_4digit());
printf("Random answer %d\n", get_random_4digit());
printf("Random answer %d\n", get_random_4digit());
Instead of testing each generated code for a disqualifying zero just generate a code without zero in it:
int generate_zero_free_code()
{
int n;
int result = 0;
for (n = 0; n < 4; n ++)
result = 10 * result + rand() % 9; // add a digit 0..8
result += 1111; // shift each digit from range 0..8 to 1..9
return result;
}
You can run the number, dividing it by 10 and checking the rest of it by 10:
int a = n // save the original value
while(a%10 != 0){
a = a / 10;
}
And then check the result:
if (a%10 != 0) printf("%d\n", n);
Edit: making it a stand alone function:
bool unique_4digits(int n)
{
while(n%10 != 0){
n = n / 10;
}
return n != 0;
}
Usage: if (unique_4digits(n)) printf("%d\n", n);
To test if the number doesn't contain any zero you can use a function that returns zero if it fails and the number if it passes the test :
bool FourDigitsWithoutZero() {
int n = get_random_4digit();
if (n % 1000 < 100 || n % 100 < 10 || n % 10 == 0) return 0;
else return n;
}
"I need not help with the function get_random_4digits because it is correct."
Actually the following does not compile,
int get_random_4digit(){
int lower = 1000, upper = 9999,answer;
answer = (rand()%(upper-lower)1)+lower;
return answer;
}
The following includes modifications that do compile, but still does not match your stated objectives::
int get_random_4digit(){
srand(clock());
int lower = 1000, upper = 9999,answer;
int range = upper-lower;
answer = lower + rand()%range;
return answer;
}
" I want to make a function that when it is called will generate a number between 1111 to 9999,"
This will do it using a helper function to test for zero:
int main(void)
{
printf( "Random answer %d\n", random_range(1111, 9999));
printf( "Random answer %d\n", random_range(1111, 9999));
printf( "Random answer %d\n", random_range(1111, 9999));
printf( "Random answer %d\n", random_range(1111, 9999));
return 0;
}
Function that does work follows:
int random_range(int min, int max)
{
bool zero = true;
char buf[10] = {0};
int res = 0;
srand(clock());
while(zero)
{
res = min + rand() % (max+1 - min);
sprintf(buf, "%d", res);
zero = if_zero(buf);
}
return res;
}
bool if_zero(const char *num)
{
while(*num)
{
if(*num == '0') return true;
num++;
}
return false;
}
This is a program to find the largest even number and its times of occurring from an input file and output it to an output file. I'm having a problem with the output, there seems to be an extra iteration that messes things up.
int main(int argc, char const *argv[])
{
int n, num, i, even, count;
FILE * fptr;
FILE * fptro;
fptr =fopen("maximpar.in", "r");
fptro=fopen("maximpar.out", "w");
/*scanning the first line from the file to get n for for()*/
fscanf(fptr, "%d", &n);
count = 0;
even = INT_MIN;
for(i = 0; i < n; i++)
{
fscanf(fptr, "%d", &num);
if( (num % 2 == 0 && num > even) || (even == num) )
/*checking for largest even number,
not sure about the ..||(even == num) part of the condition*/
{
even = num;
count++;
}
}
fprintf(fptro, "%d %d", even, count);
fclose(fptr);
fclose(fptro);
return 0;
}
Input file
6
9 6 9 8 9 8
Output file
8 3
Why isn't the output file like this? I don't understand
8 2
You need to reset your count whenever you get a new larger number.
I didn't test this, but it should work:
cate = 0;
par = INT_MIN;
for (i = 0; i < n; i++) {
fscanf(fptr, "%d", &b);
// skip odd numbers
if ((b % 2) != 0)
continue;
// get new larger number
if (b > par) {
par = b;
cate = 1;
continue;
}
// increment count on existing largest number
if (b == par)
++cate;
}
UPDATE:
I dont understand why skip iterations explicitly instead of only picking out the iterations that matter? Is there some sort of advantage?
Yes, it's better style. It allows simple single level indented if statements that can have their own comments.
It avoids a messy compound if or a triple level if/else ladder.
IMO, it's a common misconception [particularly among beginning C programmers] that a complex if will execute faster [or is somehow "better"] than several simple ones.
The first if could be thought of a "skip this iteration" test. Here, there's only one. But, for more complex code, there might be several.
The multiple condtion escapes could be handled in a single if with if (c1 || c2 || c2 || ... || c10) continue; but that gets messy fast.
Herein, for properly indented if/else ladder logic, we'd need:
if (cond1)
do_stuff1;
else
if (cond2)
do_stuff2;
else
if (cond3)
do_stuff3;
If we're not in a loop, here's a "trick" to avoid if/else ladder logic, by using do { ... } while (0);:
do {
if (cond1) {
do_stuff1;
break;
}
if (cond2) {
do_stuff2;
break;
}
if (cond3) {
do_stuff3;
break;
}
} while (0);
enclose the condition
if( ( ...&&...) ||(....) )
The answer is because count was incremented from 0 to 1 when b = 6. 2 iterations later, b = 8 and now count = 2, and 2 iterations after that, b = 8 and count = 3.
I also recommend you nest your if statement in parentheses for readability. Commenting would help too :) I'm a stats guy, and I have no idea what you are doing based on your variables' names.
You need to reset your counter inside the if block if b > par.
Like:
if( num % 2 == 0 && num >= even) {
if (num > even){
even = num;
count = 1;
} else {
count++;
}
}
Thanks.
JK
I've been at this problem for a day now and it feels like I'm getting nowhere.
What I want to do:
Generate all possible combinations of a nine-digit number between 1-9, but no digits can be the same. In other words, my goal is to generate exactly 362880 (9!) numbers, each one unique from one another and each number must contain only one of each digits. There should be no randomness involved.
What I want:
123456789
213796485
What I DON'T want:
111111111
113456789
What I've tried:
I start by creating an array to store the digits.
float num[9];
Using the principle that I num[0] can be any of the 9 digits, and num[8] has to be the one remaining, I tried nesting loops. I'll post the code, but there's no need to point out why it doesn't work because I already know why. However, I don't know how to fix it.
for (int a = 1; a < 10; a++) {
num[0] = a;
for (int b = 1; b < 9; b++) {
if (b != a)
num[1] = b;
// The code in between follows the same pattern
for (int i = 1; i < 2; i++) {
if (i != a && i != b && i != c && i != d && i != e && i != f && i != g && i != h) {
num[8] = i;
}
}
}
}
So as you can see, the last digit will always be 1, the second digit can never be 9 and so on.
So what options do I have? I tried making it so that it loops a total of 9^9 times, which would fix the problem I mentioned, but that's of course way too inefficient (and it didn't quite work as intended either).
Any ideas? I feel like it should be an easy thing to solve but I can't seem to be able to wrap my head around it.
Here is a simple solution that generates the 362880 permutations in lexicographical order:
#include <stdio.h>
#include <string.h>
void perm9(char *dest, int i) {
if (i == 9) {
printf("%.9s\n", dest);
} else {
for (char c = '1'; c <= '9'; c++) {
if (memchr(dest, c, i) == NULL) {
dest[i] = c;
perm9(dest, i + 1);
}
}
}
}
int main(void) {
char dest[9];
perm9(dest, 0);
return 0;
}
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int myatoi(const char* string) {
int i = 0;
while (*string) {
i = (i << 3) + (i<<1) + (*string -'0');
string++;
}
return i;
}
void decimal2binary(char *decimal, int *binary) {
decimal = malloc(sizeof(char) * 32);
long int dec = myatoi(decimal);
long int fraction;
long int remainder;
long int factor = 1;
long int fractionfactor = .1;
long int wholenum;
long int bin;
long int onechecker;
wholenum = (int) dec;
fraction = dec - wholenum;
while (wholenum != 0 ) {
remainder = wholenum % 2; // get remainder
bin = bin + remainder * factor; // store the binary as you get remainder
wholenum /= 2; // divide by 2
factor *= 10; // times by 10 so it goes to the next digit
}
long int binaryfrac = 0;
int i;
for (i = 0; i < 10; i++) {
fraction *= 2; // times by two first
onechecker = fraction; // onechecker is for checking if greater than one
binaryfrac += fractionfactor * onechecker; // store into binary as you go
if (onechecker == 1) {
fraction -= onechecker; // if greater than 1 subtract the 1
}
fractionfactor /= 10;
}
bin += binaryfrac;
*binary = bin;
free(decimal);
}
int main(int argc, char **argv) {
char *data;
data = malloc(sizeof(char) * 32);
int datai = 1;
if (argc != 4) {
printf("invalid number of arguments\n");
return 1;
}
if (strcmp(argv[1], "-d")) {
if (strcmp(argv[3], "-b")) {
decimal2binary(argv[2], &datai);
printf("output is : %d" , datai);
} else {
printf("invalid parameter");
}
} else {
printf("invalid parameter");
}
free(data);
return 0;
}
In this problem, myatoi works fine and the decimal2binary algorithm is correct, but every time I run the code it gives my output as 0. I do not know why. Is it a problem with pointers? I already set the address of variable data but the output still doesn't change.
./dec2bin "-d" "23" "-b"
The line:
long int fractionfactor = .1;
will set fractionfactor to 0 because the variable is defined as an integer. Try using a float or double instead.
Similarly,
long int dec = myatoi(decimal);
stores an integer value, so wholenum is unnecessary.
Instead of
i = (i << 3) + (i<<1) + (*string -'0');
the code will be much more readable as
i = i * 10 + (*string - '0');
and, with today's optimizing compilers, both versions will likely generate the same object code. In general, especially when your code isn't working, favor readability over optimization.
fraction *= 2; // times by two first
Comments like this, that simply translate code to English, are unnecessary unless you're using the language in an unusual way. You can assume the reader is familiar with the language; it's far more helpful to explain your reasoning instead.
Another coding tip: instead of writing
if (strcmp(argv[1], "-d")) {
if (strcmp(argv[3], "-b")) {
decimal2binary(argv[2], &datai);
printf("output is : %d" , datai);
} else {
printf("invalid parameter");
}
} else {
printf("invalid parameter");
}
you can refactor the nested if blocks to make them simpler and easier to understand. In general it's a good idea to check for error conditions early, to separate the error-checking from the core processing, and to explain errors as specifically as possible so the user will know how to correct them.
If you do this, it may also be easier to realize that both of the original conditions should be negated:
if (strcmp(argv[1], "-d") != 0) {
printf("Error: first parameter must be -d\n");
else if (strcmp(argv[3], "-b") != 0) {
printf("Error: third parameter must be -b\n");
} else {
decimal2binary(argv[2], &datai);
printf("Output is: %d\n" , datai);
}
void decimal2binary(char *decimal, int *binary) {
decimal = malloc(sizeof(char) * 32);
...
}
The above lines of code allocate a new block of memory to decimal, which will then no longer point to the input data. Then the line
long int dec = myatoi(decimal);
assigns the (random values in the) newly-allocated memory to dec.
So remove the line
decimal = malloc(sizeof(char) * 32);
and you will get the correct answer.
if(!strcmp(argv[3] , "-b"))
if(!strcmp(argv[3] , "-d"))
The result of the string compare function should be negated so that you can proceed. Else it will print invalid parameter. Because the strcmp returns '0' when the string is equal.
In the 'decimal2binary' function you are allocating a new memory block inside the function for the input parameter 'decimal',
decimal = malloc(sizeof(char) * 32);
This would actually overwrite your input parameter data.
UVA problem 100 - The 3n + 1 problem
I have tried all the test cases and no problems are found.
The test cases I checked:
1 10 20
100 200 125
201 210 89
900 1000 174
1000 900 174
999999 999990 259
But why I get wrong answer all the time?
here is my code:
#include "stdio.h"
unsigned long int cycle = 0, final = 0;
unsigned long int calculate(unsigned long int n)
{
if (n == 1)
{
return cycle + 1;
}
else
{
if (n % 2 == 0)
{
n = n / 2;
cycle = cycle + 1;
calculate(n);
}
else
{
n = 3 * n;
n = n + 1;
cycle = cycle+1;
calculate(n);
}
}
}
int main()
{
unsigned long int i = 0, j = 0, loop = 0;
while(scanf("%ld %ld", &i, &j) != EOF)
{
if (i > j)
{
unsigned long int t = i;
i = j;
j = t;
}
for (loop = i; loop <= j; loop++)
{
cycle = 0;
cycle = calculate(loop);
if(cycle > final)
{
final = cycle;
}
}
printf("%ld %ld %ld\n", i, j, final);
final = 0;
}
return 0;
}
The clue is that you receive i, j but it does not say that i < j for all the cases, check for that condition in your code and remember to always print in order:
<i>[space]<j>[space]<count>
If the input is "out of order" you swap the numbers even in the output, when it is clearly stated you should keep the input order.
Don't see how you're test cases actually ever worked; your recursive cases never return anything.
Here's a one liner just for reference
int three_n_plus_1(int n)
{
return n == 1 ? 1 : three_n_plus_1((n % 2 == 0) ? (n/2) : (3*n+1))+1;
}
Not quite sure how your code would work as you toast "cycle" right after calculating it because 'calculate' doesn't have explicit return values for many of its cases ( you should of had compiler warnings to that effect). if you didn't do cycle= of the cycle=calculate( then it might work?
and tying it all together :-
int three_n_plus_1(int n)
{
return n == 1 ? 1 : three_n_plus_1((n % 2 == 0) ? (n/2) : (3*n+1))+1;
}
int max_int(int a, int b) { return (a > b) ? a : b; }
int min_int(int a, int b) { return (a < b) ? a : b; }
int main(int argc, char* argv[])
{
int i,j;
while(scanf("%d %d",&i, &j) == 2)
{
int value, largest_cycle = 0, last = max_int(i,j);
for(value = min_int(i,j); value <= last; value++) largest_cycle = max_int(largest_cycle, three_n_plus_1(value));
printf("%d %d %d\r\n",i, j, largest_cycle);
}
}
Part 1
This is the hailstone sequence, right? You're trying to determine the length of the hailstone sequence starting from a given N. You know, you really should take out that ugly global variable. It's trivial to calculate it recursively:
long int hailstone_sequence_length(long int n)
{
if (n == 1) {
return 1;
} else if (n % 2 == 0) {
return hailstone_sequence_length(n / 2) + 1;
} else {
return hailstone_sequence_length(3*n + 1) + 1;
}
}
Notice how the cycle variable is gone. It is unnecessary, because each call just has to add 1 to the value computed by the recursive call. The recursion bottoms out at 1, and so we count that as 1. All other recursive steps add 1 to that, and so at the end we are left with the sequence length.
Careful: this approach requires a stack depth proportional to the input n.
I dropped the use of unsigned because it's an inappropriate type for doing most math. When you subtract 1 from (unsigned long) 0, you get a large positive number that is one less than a power of two. This is not a sane behavior in most situations (but exactly the right one in a few).
Now let's discuss where you went wrong. Your original code attempts to measure the hailstone sequence length by modifying a global counter called cycle. However, the main function expects calculate to return a value: you have cycle = calculate(...).
The problem is that two of your cases do not return anything! It is undefined behavior to extract a return value from a function that didn't return anything.
The (n == 1) case does return something but it also has a bug: it fails to increment cycle; it just returns cycle + 1, leaving cycle with the original value.
Part 2
Looking at the main. Let's reformat it a little bit.
int main()
{
unsigned long int i=0,j=0,loop=0;
Change these to long. By the way %ld in scanf expects long anyway, not unsigned long.
while (scanf("%ld %ld",&i,&j) != EOF)
Be careful with scanf: it has more return values than just EOF. Scanf will return EOF if it is not able to make a conversion. If it is able to scan one number, but not the second one, it will return 1. Basically a better test here is != 2. If scanf does not return two, something went wrong with the input.
{
if(i > j)
{
unsigned long int t=i;i=j;j=t;
}
for(loop=i;loop<=j;loop++)
{
cycle=0;
cycle=calculate(loop );
if(cycle>final)
{
final=cycle;
}
}
calculate is called hailstone_sequence_length now, and so this block can just have a local variable: { long len = hailstone_sequence_length(loop); if (len > final) final = len; }
Maybe final should be called max_length?
printf("%ld %ld %ld\n",i,j,final);
final=0;
final should be a local variable in this loop since it is separately used for each test case. Then you don't have to remember to set it to 0.
}
return 0;
}