The program should read 5 numbers from the user. Then it has to draw 5 'lines' according to the rate of this numbers. At the biggest number it has to draw 20 ' * '-s and for every other number it has to draw less ' * '-s in proportion to the biggest number.
There's a program which can check if my code is correct, and I tried a lot of things to make it work, but I always got bad result (right side on the picture). Then I was checking the solution of similar exercises and I found that the only difference is that, in the soutions there's a dot after the defined number
#define SEGMENT 20.
And when I changed just this one thing in my code it worked correctly (left side on the picture). Can someone explain why does it make such a huge difference and why it's working without dot too with some inputs?
Here's the full code:
#include <stdio.h>
#define PCS 5
#define SEGMENT 20
//The only difference in the line above
//#define SEGMENT 20.
int main(void) {
double nums[PCS];
int max;
//Reding the five numbers and finding the biggest value
for(int i = 0; i < PCS; i++) {
scanf("%lf", &nums[i]);
if(i == 0 || max < nums[i]) {
max = nums[i];
}
}
//Calcualte the rate for the lines
double rate = SEGMENT / max;
//Print the lines
for(int i = 0; i < PCS; i++) {
for(int j = 0; j < (int)(nums[i]*rate); j++) {
printf("*");
}
printf("\n");
}
return 0;
}
In context of this statement
double rate = SEGMENT / max;
20 is an integer. Hence the expression 20 / max is also an integer with remainder dropped before being assigned to rate
20. is the same as 20.0 - it's a floating point real value. Hence, 20.0 / max gets evaluated as a real number (with max getting effectively promoted from integer to float before the division operation is applied). And the result of that expression is directly assigned to rate without any rounding.
If it helps, you can avoid the floating point computation (and all the weirdness that goes with fp) by doing integer math and having the division operation performed as the last step of your calculation.
That is, instead of this:
#define SEGMENT 20.0
...
double rate = SEGMENT / max;
...
for(int j = 0; j < (int)(nums[i]*rate); j++) {
This:
#define SEGMENT 20
...
for(int j = 0; j < (int)(nums[i]*SEGMENT)/max; j++) {
This keeps your math in the integer space and should still work. (Disclaimer: nums is still an array of double, so the expression (nums[i]*20)/max is still a floating point value, but it's not going to incur a rounding error.
When you divide an int by an int you get int (truncated);
7 / 2 == 3
7 / 2.0 == 3.5 (as well as 7.0 / 2)
In you code 20. is just a shorthand for 20.0
Related
Goodevening everyone, i'm learning now to loop an array of numbers, and i'm just wondering,
how can i square the numbers in my array depending on the input of how many times i should square it
int main() {
int nums[5] = {1, 2, 3, 4, 5};
int loop;
int a;`enter code here`
int pro,pro1,pro2,pro3,pro4;
int spacing = 3;
int i = 0
scanf("%d", &a);
do{
pro = pow(nums[0],2);
pro1 = pow(nums[1],2);
pro2 = pow(nums[2],2);
pro3 = pow(nums[3],2);
pro4 = pow(nums[4],2);
i++
}while (i != a);
printf("%0*d\n", spacing, pro);
printf("%0*d\n", spacing, pro1);
printf("%0*d\n", spacing, pro2);
printf("%0*d\n", spacing, pro3);
printf("%0*d\n", spacing, pro4);
return 0;
}
this is my code so far, i wanted to loop it, and get the results like this:
001
016
081
256
625
or like this
001
256
6561
65536
390625
but i always get this:
001
004
009
016
025
please help me understand thank you
There is a strong error and a number of possible improvements in your code.
The error has already be identified by #SylvainChaugny and is that you re-use original nums value on each and every iteration. The improvements are:
you are using 5 variables pro to pro5 and process them the same. Better to make an array for them, or even better re-use the nums array.
if you want to later extend your program to 6 values, you would have to consistently look through the code to search what needs to be changed: better to use a constant or as you have a literal initialization ask the size to the compiler
you are using pow to process integer values. This is not efficient because as pow takes and returns double values you force a conversion from int to double and back. In addition it might be dangerous for large values: a double has less precision than an int on 64 bits architectures (48 mantissa bits for a double, 64 bits for an int). So it can lead to incorrect results.
So your code could become:
#include <stdio.h>
int main() {
int nums[] = {1, 2, 3, 4, 5};
int len = sizeof (nums) / sizeof (*nums); // idiomatic way for the length of an array
int a;
int spacing = 3;
scanf("%d", &a);
for (int i=0; i<a; i++) {
for (int j=0; j<len; j++) {
nums[j] = nums[j] * nums[j];
}
}
for (int i=0; i<len; i++) {
printf("%0*d\n", spacing, nums[i]);
}
return 0;
}
It is shorter to type (not only laziness but also less sensitive to typos) and it gives the expected result ;-). In addition, if you want to add a value to the array, it can be done be changing one single line.
Your problem is that you always reassign your pro variables to the exact same value (pow(nums[0], 2), pow(nuws[1], 2), ....
So no matter what the value of a is, you always will have your pro3 variable to be equal to pow(nums[3], 2).
You first have to assign the initial values to your pro variables, then use them in the calls to pow(), to be able to square your previous result.
I'm new to competitive programming and I participated in Codeforces #653 in which I spent the whole time solving this problem and did not submit any code because of exceeded time limits or wrong answer for test cases.
I wanted to solve this problem - https://codeforces.com/contest/1374/problem/A
You are given three integers x,y and n. Your task is to find the maximum integer k such that 0 â¤k⤠n that k mod x=y, where mod is modulo operation. Many programming languages use percent operator % to implement it.
In other words, with given x,y and n you need to find the maximum possible integer from 0 to n that has the remainder y modulo x
You have to answer t independent test cases. It is guaranteed that such k exists for each test case.
I wrote this following code:
#include <stdio.h>
int main(){
int i,t,j,k=0;
int x[60000],y[60000],n[60000],prev[60000];
scanf("%d",&t);
for(i=0; i<t; i++){
scanf("%d %d %d",&x[i],&y[i],&n[i]);
}
for(i=0; i<t; i++){
for(j=0, k=0; j<n[i]; j++ , k++){
if(j%x[i]==y[i]){
prev[i]=k;
}
}
}
for(i=0; i<t; i++){
printf("%d",prev[i]);
printf("\n");
}
return 0;
}
Everything was working fine but for some test cases I'm getting different answers.
This was the expected output
12339
0
15
54306
999999995
185
999999998
and my output was this:
12339
0
5
54306
999999995
185
999999998
I did not understand why I got 5 instead of 15 keeping all the other outputs correct and also could anyone please help me to optimize the code, its taking too long to compile for large inputs.
For the 1st part of your question, why the answer is wrong - has been answered nicely by others already. For the 2nd part about efficiency, the solution doesn't need any extra loop except the loop for iterating over the test case.
The solution could be as easy as this:
k = n - ((n - y) % x)
For example:x = 7, y = 5, n = 12345. Then,
k = 12345 - ((12345 - 5) % 7)
= 12339
This small piece of code could get you accepted:
#include <stdio.h>
int main()
{
int t, x, y, n;
scanf("%d", &t);
while (t > 0) {
scanf("%d %d %d", &x, &y, &n);
printf("%d\n", n - ((n - y) % x));
t--;
}
}
The reason you were getting TLE was because your code was taking too long. As I can see n could be upto 10^9 and so a O(N) solution would easily time-out at such constraints. Add to that, the fact that your code would be given upto 5*10^4 test cases. So, for your code to work, it should be much faster than O(N) time complexity. I have explained a better approach below which would satisfy the given constraints.
Optimised Approach :
For each test case, we are given x, y, n. And we have to find the largest number( let's say ans) between 0 to n such that ans%x = y.
Let's first find the remainder when we divide n by x. So, remainder = n%x. Now, if the remainder >= y, this means that we would have to reduce n such that it will leave a smaller remainder that is => a remainder equal to y. For this, we can simply reduce n by (remainder - y) amount.
Example :
For better understanding, lets see an example where
x = 10, y = 5, n = 16.
In this case remainder = n%x = 6. Now remainder > y, so we can just reduce our n by (remainder - y), that is n now becomes 15. We see that 15%x = y and so that's our answer.
The other case we might get is remainder < y. In this case, we have to increase the remainder. But we can't increase n (since it is the upper limit). So, what we can instead do is subtract x from n. The remainder will still be same. But now we are allowed to increase n by some amount which results in remainder to be y. For that we simply increase n by an amount y - remainder, so that new remainder will be equal to y.
Example :
Let's consider example where
x = 10, y = 5 and n = 24
Here, remainder = n%x = 4. So remainder < y. We subtract x from n, so n becomes 14 but still n%x = 4, so remainder remains same. But, we now have the advantage that we can increase x so that remainder would be equal to y. Since our remainder is 1 less than required (y), we increase n by 1 (or y - remainder = 1). Thus, we have the answer = 15.
This approach has time complexity of O(1) which is far better than O(N) approach. I hope that you got the idea of the approach I was trying to explain. Below is the pseudocode for same -
Pseudocode
remainder = n%x
if (remainder >= y) // Case 1
return n - ( remainder - y ) as the answer
else // Case 2
return ( n - x ) + ( y - remainder ) as the answer
Hope this helps !
The first part of this answer will explain what's wrong in your logic. The second part will contain a plot twist.
How to get the correct answer
As you have correctly been told in comments by #ErichKitzmueller your inner loop searches k values in the range [0-n[. In other words, if you are not familiar to my mathematical notation, you are not even considering the value n that is not included in your loop search, as you do for(j=0, k=0; j<n[i]; j++ , k++).
For the record, [0-n[ means "range from to 0 to n including 0 and not including n.
If you have to search the maximum value satisfying a given requirement... why starting counting from 0? You just need starting from the right limit of the range and loop backwards. The first k you will find satisfying the condition will be your output, so you'll just need to save it and exit the inner loop.
No need to find ALL the numbers satisfying the condition and overwrite them until the last is found (as you do).
The main loop of your solution would become something like that:
for(i=0; i<t; i++){
int found = 0;
for(k=n[i]; k>=0 && found==0; k--)
{
if( ( k % x[i] ) == y[i] )
{
prev[i] = k;
found = 1;
}
}
}
The plot twist (the REAL solution)
The previous solution will lead to correct answers... anyway it will be rejected as it exceeds the time limit.
Actually, all these competitive coding problems are all based on asking for a problem that in some way is simpler than it looks. In other words, it's always possible to find a way (usually after a mathematical analysis) that have a lower computational complexity than the one of the first solution that comes to your mind.
In this case we have to think:
What is the reminder of a division? y = k%x = k - x*int(k/x)
When has this expression its max? When k=n. So y = k - x*int(n/x)
So k = x*int(n/x) + y
Finally, we want make sure that this number is lower than n. If it is, we subtract x
The code becomes something like this:
#include <stdio.h>
int main(){
int i, t;
int x[60000],y[60000],n[60000],k[60000];
scanf("%d",&t);
for(i=0; i<t; i++){
scanf("%d %d %d",&x[i],&y[i],&n[i]);
}
for(i=0; i<t; i++){
int div = n[i] / x[i]; // Since div is an integer, only the integer part of the division is stored to div
k[i] = x[i] * div + y[i];
if( k[i] > n[i] )
k[i] -= x[i];
}
for(i=0; i<t; i++){
printf("%d", k[i]);
printf("\n");
}
return 0;
}
I've tested the solution on Codeforce, and it has been accepted.
the following proposed code:
cleanly compiles
performs the desired functionality
is very quick (could be made quicker via different I/O functions)
and now, the proposed code:
#include <stdio.h>
int main()
{
int x;
int y;
int n;
size_t t;
scanf("%zu",&t);
for( ; t; t-- )
{
scanf( "%d %d %d", &x, &y, &n );
printf( "%d\n", n - ((n - y) % x) );
}
return 0;
}
I am trying to simulate the propagation of a worm across a network made of 100,000 computers. The simulation itself is very simple and I don't need any help except that for some reason, I am only getting every third random number.
Only the computers whose index modulo 1000 is less than 10 can be infected so when 1000 computers are infected, the program should be done. For some reason, my program only gets 329. When I lower the goal number and check the contents of the array, only every third computer has been changed and it is a consistent pattern. For example at the end of the array, only computers 98001, 98004, 98007, 99002, 99005, 99008 are changed even though the computers in between (98002, 98003, etc.) should be changed as well. The pattern holds all the way to the beginning of the array. When I try to get all 1000 changed, the program goes into an infinite loop and is stuck at 329.
Edit: I just discovered that if I lower the NETSIZE to 10,000 and the goal in the while loop to 100, it doesn't skip anything. Does that mean the problem has something to do with a rounding error? Someone who knows more about C than me must know the answer.
Thanks.
#include <stdio.h>
#include <stdlib.h>
#define NETSIZE 100000
double rand01();
void initNetwork();
unsigned char network[NETSIZE];
int scanrate = 3;
int infectedCount;
int scans;
int ind;
int time;
int main(void) {
initNetwork();
time = 0;
infectedCount = 1;
while (infectedCount < 1000) { //changing 1000 to 329 stops the infinite loop
scans = infectedCount * scanrate;
for (int j = 0; j < scans; j++) {
ind = (int) (rand01() * NETSIZE);
if (network[ind] == 0) {
network[ind] = 1;
infectedCount++;
}
}
time++;
}
for (int k = 0; k < NETSIZE; k++) {
if (network[k] == 1) printf("%d at %d\n", network[k], k);
}
}
double rand01() {
double temp;
temp = (rand() + 0.1) / (RAND_MAX + 1.0);
return temp;
}
void initNetwork() {
for (int i = 0; i < NETSIZE; i++) {
if (i % 1000 < 10) {
network[i] = 0;
} else {
network[i] = 2;
}
}
network[1000] = 1;
}
In the above code, I expect the code to run until the 1000 vulnerable indexes are changed from 0 to 1.
Converting comments into an answer.
What is RAND_MAX on your system? If it is a 15-bit or 16-bit value, you probably aren't getting good enough quantization when converted to double. If it is a 31-bit or bigger number, that (probably) won't be the issue. You need to investigate what values are generated by just the rand01() function with different seeds, plus the multiplication and cast to integer â simply print the results and sort -n | uniq -c to see how uniform the results are.
On my system RAND_MAX is only 32767. Do you think that might be why my results might not be granular enough? Now that you've made me think about it, there would only be 32,767 possible values and my network array is 100,000 possible values. Which corresponds about about the 1/3 results I am getting.
Yes, I think that is very probably the problem. You want 100,000 different values, but your random number generator can only generate about 33,000 different values, which is awfully close to your 1:3 metric. It also explains immediately why you got good results when you reduced the multiplier from 100,000 to 10,000.
You could try:
double rand01(void)
{
assert(RAND_MAX == 32767);
return ((rand() << 15) + rand()) / ((RAND_MAX + 1.0) * (RAND_MAX + 1.0));
}
Or you could use an alternative random number generator â for example, POSIX defines both the drand48() family of functions and
random(), with corresponding seed-setting functions where needed.
Yeah, the problem I am having is that the RAND_MAX value on my system is only 32767 and I am trying to effectively spread that out over 100,000 values which results in about only every third number ever showing up.
In my defense, the person who suggested the rand01() function has a PhD in Computer Science, but I think he ran this code on our school's main computer which probably has a much bigger RAND_MAX value.
#JonathanLeffler deserves credit for this solution.
I'm new to C programming (I have some very basic experience with programming via vb.NET), and I'm attempting to write a program for the Project Euler Problem #1.
https://projecteuler.net/problem=1
Algorithm
The challenge requires the programmer to find the sum of all multiples of 3 or 5 (inclusive) below 1000 (I used intInput to allow the user to enter an integer in place of 1000).
My current solution takes the input, and decrements it by 1 until (intInput - n) % 3 = 0, that is, until the next nearest multiple of 3 under the input integer is found.
The program then cycles through all integers from 1 to ((intInput - n) / 3), adding each integer to the sum of the previous integers, so long as the current integer is not a multiple of 5, in which case, it is skipped.
The resultant sum is then stored in intThreeMultiplier.
The above process is then repeated, using 5 in place of 3 to find the highest multiple of 5 under intInput, and then cycles through integers 1 to ((intInput - n) / 5), not skipping multiples of 3 this time, and stores the sum in intFiveMultiplier.
The output sum is then calculated via sum = (3 * intThreeMultiplier) + (5 * intFiveMultiplier).
The Problem
Whenever I compile and run my code, the user is allowed to input an integer, and then the program crashes. I have determined that the cause has something to do with the first For loop, but I can't figure out what it is.
I have commented out everything following the offending code fragment.
Source Code:
#include <stdio.h>
#include <stdlib.h>
void main()
{
int intInput = 0; /*Holds the target number (1000 in the challenge statement.)*/
int n = 0;
int count = 0;
int intThreeMultiplier = 1;
int intFiveMultiplier = 1;
printf("Please enter a positive integer.\n");
scanf("%d",intInput);
for( ; (((intInput - n) % 3) != 0) ; n++)
{}
/*for(; count <= ((intInput - n) / 3); count++)
{
if ((count % 5) != 0)
{
intThreeMultiplier += count;
}
}
count = 0;
for(n = 0 ; ((intInput - n) % 5) != 0 ; n++)
{}
for(; count <= ((intInput - n) / 5) ; count++)
{
intFiveMultiplier += count;
}
int sum = (3 * intThreeMultiplier) + (5 * intFiveMultiplier);
printf("The sume of all multiples of 3 or 5 (inclusively) under %d is %d.",intInput, sum);*/
}
This is my first time posting on StackOverflow, so I apologize in advance if I have broken any of the rules for asking questions, and would appreciate any feedback with respect to this.
In addition, I am extremely open to any suggestions regarding coding practices, or any rookie mistakes I've made with C.
Thanks!
scanf("%d",intInput);
might be
scanf("%d", &intInput); // note the ampersand
scanf need the address the variable where the content is to be stored. Why scanf must take the address of operator
For debugging only, print the input to verify that the input is accepted correctly, something like
printf("intInput = %d\n", intInput);
The first thing you need when you are inputting intInput you should use:
scanf("%d", &intInput);
Because scanf() need as an argument of a pointer to your variable. You are doing this by just putting the & sign before your int.
In addition I think that you should double check your algorithm, because you are summing up some numbers more than once. :)
The problem I am given is the following:
Write a program to discover the answer to this puzzle:"Let's say men and women are paid equally (from the same uniform distribution). If women date randomly and marry the first man with a higher salary, what fraction of the population will get married?"
From this site
My issue is that it seems that the percent married figure I am getting is wrong. Another poster asked this same question on the programmers exchange before, and the percentage getting married should be ~68%. However, I am getting closer to 75% (with a lot of variance). If anyone can take a look and let me know where I went wrong, I would be very grateful.
I realize, looking at the other question that was on the programmers exchange, that this is not the most efficient way to solve the problem. However, I would like to solve the problem in this manner before using more efficient approaches.
My code is below, the bulk of the problem is "solved" in the test function:
#include <cs50.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define ARRAY_SIZE 100
#define MARRIED 1
#define SINGLE 0
#define MAX_SALARY 1000000
bool arrayContains(int* array, int val);
int test();
int main()
{
printf("Trial count: ");
int trials = GetInt();
int sum = 0;
for(int i = 0; i < trials; i++)
{
sum += test();
}
int average = (sum/trials) * 100;
printf("Approximately %d %% of the population will get married\n", average / ARRAY_SIZE);
}
int test()
{
srand(time(NULL));
int femArray[ARRAY_SIZE][2];
int maleArray[ARRAY_SIZE][2];
// load up random numbers
for (int i = 0; i < ARRAY_SIZE; i++)
{
femArray[i][0] = (rand() % MAX_SALARY);
femArray[i][1] = SINGLE;
maleArray[i][0] = (rand() % MAX_SALARY);
maleArray[i][1] = SINGLE;
}
srand(time(NULL));
int singleFemales = 0;
for (int k = 0; k < ARRAY_SIZE; k++)
{
int searches = 0; // count the unsuccessful matches
int checkedMates[ARRAY_SIZE] = {[0 ... ARRAY_SIZE - 1] = ARRAY_SIZE + 1};
while(true)
{
// ARRAY_SIZE - k is number of available people, subtract searches for people left
// checked all possible mates
if(((ARRAY_SIZE - k) - searches) == 0)
{
singleFemales++;
break;
}
int randMale = rand() % ARRAY_SIZE; // find a random male
while(arrayContains(checkedMates, randMale)) // ensure that the male was not checked earlier
{
randMale = rand() % ARRAY_SIZE;
}
checkedMates[searches] = randMale;
// male has a greater income and is single
if((femArray[k][0] < maleArray[randMale][0]) && (maleArray[randMale][1] == SINGLE))
{
femArray[k][1] = MARRIED;
maleArray[randMale][1] = MARRIED;
break;
}
else
{
searches++;
continue;
}
}
}
return ARRAY_SIZE - singleFemales;
}
bool arrayContains(int* array, int val)
{
for(int i = 0; i < ARRAY_SIZE; i++)
{
if (array[i] == val)
return true;
}
return false;
}
In the first place, there is some ambiguity in the problem as to what it means for the women to "date randomly". There are at least two plausible interpretations:
You cycle through the unmarried women, with each one randomly drawing one of the unmarried men and deciding, based on salary, whether to marry. On each pass through the available women, this probably results in some available men being dated by multiple women, and others being dated by none.
You divide each trial into rounds. In each round, you randomly shuffle the unmarried men among the unmarried women, so that each unmarried man dates exactly one unmarried woman.
In either case, you must repeat the matching until there are no more matches possible, which occurs when the maximum salary among eligible men is less than or equal to the minimum salary among eligible women.
In my tests, the two interpretations produced slightly different statistics: about 69.5% married using interpretation 1, and about 67.6% using interpretation 2. 100 trials of 100 potential couples each was enough to produce fairly low variance between runs. In the common (non-statistical) sense of the term, for example, the results from one set of 10 runs varied between 67.13% and 68.27%.
You appear not to take either of those interpretations, however. If I'm reading your code correctly, you go through the women exactly once, and for each one you keep drawing random men until either you find one that that woman can marry or you have tested every one. It should be clear that this yields a greater chance for women early in the list to be married, and that order-based bias will at minimum increase the variance of your results. I find it plausible that it also exerts a net bias toward more marriages, but I don't have a good argument in support.
Additionally, as I wrote in comments, you introduce some bias through the way you select random integers. The rand() function returns an int between 0 and RAND_MAX, inclusive, for RAND_MAX + 1 possible values. For the sake of argument, let's suppose those values are uniformly distributed over that range. If you use the % operator to shrink the range of the result to N possible values, then that result is still uniformly distributed only if N evenly divides RAND_MAX + 1, because otherwise more rand() results map to some values than map to others. In fact, this applies to any strictly mathematical transformation you might think of to narrow the range of the rand() results.
For the salaries, I don't see why you even bother to map them to a restricted range. RAND_MAX is as good a maximum salary as any other; the statistics gleaned from the simulation don't depend on the range of salaries; but only on their uniform distribution.
For selecting random indices into your arrays, however, either for drawing men or for shuffling, you do need a restricted range, so you do need to take care. The best way to reduce bias in this case is to force the random numbers drawn to come from a range that is evenly divisible by the number of options by re-drawing as many times as necessary to ensure it:
/*
* Returns a random `int` in the half-open interval [0, upper_bound).
* upper_bound must be positive, and should not exceed RAND_MAX + 1.
*/
int random_draw(int upper_bound) {
/* integer division truncates the remainder: */
int rand_bound = (RAND_MAX / upper_bound) * upper_bound;
for (;;) {
int r = rand();
if (r < rand_bound) {
return r % upper_bound;
}
}
}