My professor uses this site (e-olymp.com) that automatically grades your solution in %
For this homework we have to use pointers to solve these tasks. I had a problem with this one:
The array of real numbers is given. Calculate the twice value of the minimum element in array.
Input
First line contains the number n (n ≤ 100) of elements in array. Second line contains n real numbers - the elements of array. Each value does not exceed 100 by absolute value.
Output
Print the twice value of the minimum element in array with 2 decimal digits.
My solution, works perfectly fine in compiler but gives 0%, idk where is the mistake, could you take a look at this one?
#include<stdio.h>
#include <malloc.h>
int z, x;
double fx;
int main(void){
double *c = (double *)malloc(x*sizeof(double));
scanf("%d", &x);
fx=100;
for(z=0; z<x; z++){
scanf("%lf", c+z);
if(fx>*(c+z)) fx=*(c+z);
}
printf("%.2lf", fx*2);
free(c);
return 0;
}
Related
The natural logarithm can be approximated by the following series.
If x is input through the keyboard, write a program to calculate the
sum of first seven terms of this series.
I wrote the program as:
#include <stdio.h>
#include<math.h>
int main()
{
float x, i, sum, log_sum;
printf("Enter the value of x : ");
scanf("%f",&x);
sum=0;
for (i=2; i<=7; i++)
{
sum=sum+((1/2)*pow(((x-1)/x), i));
}
log_sum=((x-1)/x)+sum;
printf("\nSum of log series is %f\n",log_sum);
return 0;
}
The output is not matching with calculator answer. What maybe wrong here?
In the for loop you multiply everything by (1/2), given 1 and 2 are integers, the answer will be 0. Try multiplying by 0.5 instead, does that solve the problem?
Had an interview today and I was asked the following question - given two arrays arr1 and arr2 of chars where they contain only numbers and one dot and also given a value m, sum them into one array of chars where they contain m digits after the dot. The program should be written in C. The algorithm was not important for them, they just gave me a compiler and 20 minutes to pass their tests.
First of all I though to find the maximum length and iterate through the array from the end and sum the values while keeping the carry:
int length = (firstLength < secondLength) ? secondLength : firstLength;
char[length] result;
for (int i = length - 1; i >= 0; i--) {
// TODO: add code
}
The problem is that for some reason I'm not sure what is the right way to perform that sum while keeping with the dot. This loop should just perform the look and not counter to k. I mean that at this point I thought just adding the values and at the end i'll insert another loop which will print k values after the dot.
My question is how should look the first loop I mentioned (the one that actually sums), I'm really got stuck on it.
The algorithm was not important
Ok, I'll let libc do it for me in that case (obviously error handling is missing):
void sum(char *as, char *bs, char *out, int precision)
{
float a, b;
sscanf(as, "%f", &a);
sscanf(bs, "%f", &b);
a += b;
sprintf(out, "%.*f", precision, a);
}
It actually took me a lot longer than 20 mins to do this. The code is fairly long too so I don't plan on posting it here. In a nutshell, the code does:
normalize the 2 numbers into 2 new strings so they have the same number of decimal digits
allocate a new string with length of longer of the 2 strings above + 1
add the 2 strings together, 2 digits at a time, with carrier
it is not clear if the final answer needs to be rounded. If not, just expand/truncate the decimals to m digits. Remove any leading zero if needed.
I am not sure whether this is the best solution or not but here's a solution and I hope it helps.
#include<stdio.h>
#include<math.h>
double convertNumber(char *arr){
int i;
int flag_d=0; //To check whether we are reading digits before or after decimal
double a=0;
int j=1;
for(i=0;i<arr[i]!='\0';i++){
if(arr[i] !='.'){
if(flag_d==0)
a = a*10 + arr[i]-48;
else{
a = a + (arr[i]-48.0)/pow(10, j);
j++;
}
}else{
flag_d=1;
}
}
return a;
}
int main() {
char num1[] = "23.20";
char num2[] = "20.2";
printf("%.6lf", convertNumber(num1) + convertNumber(num2));
}
In an interview they asked me to find out the missing number from an array.
array will be having number from 1 to N.
My Approach:
int main()
{
int ar[20];
int sum = 0;
int n;
printf("enter numb of elements\n");
scanf("%d", &n);
printf("enter array numbers\n");
for(int i = 0; i<n;i++){
scanf("%d", &ar[i]);
sum +=ar[i];
}
printf("missing num=%d", ((n*(n+1))/2)-sum);
}
But interviewer did not call back after first round of interview.
I don't know what is wrong with my approach.
Some issues with your code:
The algorithm is wrong (off by one): If the array contains all numbers from 1 to N except for one missing number, then it has N-1 elements. Your code reads N elements. (Alternatively, if the array actually has N elements, then the target sum is (N + 1) * (N + 2) / 2 (sum of numbers from 1 to N+1), not N * (N + 1) / 2.)
Includes are missing (in particular, #include <stdio.h>). That means the calls to printf / scanf have undefined behavior.
int main() should be int main(void).
None of the scanf calls check their return value. That means your code doesn't realize when reading input fails, producing garbage output.
If n is bigger than 20, your code silently writes outside the bounds of ar. That's a classic buffer overflow.
The previous point is especially unfortunate because your code doesn't even need the array. All you do with the input numbers is to add them up in sum, which doesn't require a separate array.
Your formatting is inconsistent in for(int i = 0; i<n;i++){. Why is there no space in for(int and i<n;i++){, but there are spaces around i = 0;?
Depending on how big N is, n*(n+1) could overflow.
The last line of output produced by your code is missing its terminating newline: printf("missing num=%d\n", ...);
Task: Calculate the 25 values of the function y = ax'2 + bx + c on the interval [e, f], save them in the array Y and find the minimum and maximum values in this array.
#include <stdio.h>
#include <math.h>
int main()
{
float Y[25];
int i;
int x=3,a,b,c;
double y = a*pow(x,2)+b*x+c;
printf("a = ", b);
scanf("%d", &a);
printf("b = ", a);
scanf("%d", &b);
printf("c = ", c);
scanf("%d", &c);
for(i=0;i<25;i++)
{
printf("%f",y); //output results, not needed
x++;
}
system("pause");
}
Problems:
Cant understand how can I use "interval [e,f]" here
Cant understand how to save values to array using C libraries
Cant understand how to write/make a cycle, which will find the
minimum and maximum values
Finally, dont know what exactly i need to do to solve task
You must first ask the user for the values of a, b, c or initialize those variables, and ask for the interval values of e, f, or initialize those variables.
Now you must calculate double interval= (f - e)/25.0 so you have the interval.
Then you must have a loop for (int i=0, double x=e; i<25; i++, x += interval) and calculate each value of the function. You can choose to store the result in an array (declare one at the top) or print them directly.
Problems:
Cant understand how can I use "interval [e,f]" here
(f-e) / 25(interval steps)
Cant understand how to save values to array using C libraries
You need to use some form of loop to traverse the array and save the result of your calculation at every interval step. Something like this:
for(int i = 0; i < SIZE; i++)
// SIZE in this case 25, so you traverse from 0-24 since arrays start 0
Cant understand how to write/make a cycle, which will find the minimum and maximum values
For both cases:
traverse the array with some form of loop and check every item e.g. (again) something like this: for(int i = 0; i < SIZE; i++)
For min:
Initialize a double value(key) with the first element of your array
Loop through your array searching for elements smaller than your initial key value.
if your array at position i is smaller than key, save key = array[i];
For max:
Initialize a double value(key) with 0;
Loop through your array searching for elements bigger than your initial key value.
if your array at position i is bigger than key, save key = array[i];
Finally, dont know what exactly i need to do to solve task
Initialize your variables(yourself or through user input)
Create a function that calculates a*x^2 + b*x + c n times for every step of your interval.
Create a function for min & max that loops through your array and returns the smallest/biggest value.
Thats pretty much it. I will refrain from posting code(for now), since this looks like an assignment to me and I am confident that you can write the code with the information #Paul Ogilvie & I have provided yourself. Good Luck
#include<stdio.h>
#include<math.h>
int main()
{
double y[25];
double x,a,b,c,e,f;
int i,j=0;
printf("Enter a:",&a);
scanf("%lf",&a);
printf("Enter b:",&b);
scanf("%lf",&b);
printf("Enter c:",&c);
scanf("%lf",&c);
printf("Starting Range:",&e);
scanf("%lf",&e);
printf("Ending Range:",&f);
scanf("%lf",&f);
for(i=e;i<=f;i++)
{
y[j++]=(a*pow(i,2))+(b*i)+c;
}
printf("\nThe Maximum element in the given interval is %lf",y[j-1]);
printf("\nThe Minimum element in the given interval is %lf",y[0]);
}
Good LUCK!
Check the image
This is my 1st post so have that in mind while reading my question.
I have an exam of a colloquium but my code does not provide me the correct result.
So if anyone could help me that would be great. :)
These are the informations that are provided in the exam:
A function y=f(x)=ax^2+bx+c
We have to find the surface that is below the chart but keep in mind that dx(Delta X)=B-A and the height goes like this: A,A+dx,A+2dx, .... , B-dx.
As dx value gets lower the surface will be more accurate.
You have to write the program so that the surface with precision 0.001
This is my code so could someone who is good in C check it please.
Thank you.
#include <stdio.h>
#include <math.h>
int main()
{
int a,b,c;
double A,B,dx,p,D,q,x,y,nv,nv1,nv2,sv;
do{
printf("Insert a & b: "),scanf("%lf %lf",&A,&B);
} while(A<1 || B<1);
nv=dx=B-A;
do{
printf("enter odds: "),scanf("%d %d %d",&a,&b,&c);
p=(-b)/2;
D=sqrt(pow(b,2)-4*a*c);
q= -D/4*a;
} while( a<0 || p<0 || q<0);
do{
sv=nv;
dx/=2;
nv=0;
for(x=A;x<p;x+=dx)
for(dx=B-A;dx<q;dx/=2)
nv1+=x*dx;
for(y=p;y<=B;y+=dx)
for(dx=q;dx<B;dx/=2)
nv2+=y*dx;
nv=nv1+nv2;
}while(fabs(nv-sv)>0.001);
printf("The surface is %lf",nv);
return 0;
}
You want to find the approximation of a definite integral of a quadratic function. There are several issues with your code:
What is the restriction of A ≥ 1 and B ≥ 1 for? A parabola is defined over the whole abscissa. If anything, you should enforce that the input is numeric and that two values were given.
You don't need to find the vertex of the parabola. Your task is to create small rectangles based on the left x value of each interval as the image shows. Therefore, you don't need p and q. And you shouldn't enforce that the vertex is in the first quadrant on the input without indication.
Why are the coefficients of the parabola integers? Make them doubles to be consistent.
Because you don't need to know the vertex, you don't need to split your loop in two. In your code, you don't even check that p is between A and B, which is a requirement of cour code.
What is the inner loop for? You are supposed to just calculate the area of the current rectangle here. What's worse: you re-use the variable dx as iteration variable, which means you lose it as an indicator of how large your current interval is.
The repeated incrementing of dx may lead to an accumulated floating-point error when the number of intervals is large. A common technique to avoid this is to use an integer variable for loop control and the determine the actual floating-point variable by multiplication.
The absolute value as a convergence criterion may lead to problems with small and big numbers. The iteration ends too early for small values and it may never reach the criterion for big numbers, where a difference of 0.001 cannot be resolved.
Here's a version of your code that puts all that into practice:
#include <stdio.h>
#include <math.h>
int main()
{
double a, b, c;
double A, B;
printf("Lower and upper limit A, B: ");
scanf("%lf %lf", &A, &B);
printf("enter coefficients a, b, c: ");
scanf("%lf %lf %lf", &a, &b, &c);
double nv = 0;
double sv;
int n = 1;
do {
int i;
double dx;
sv = nv;
n *= 2;
dx = (B - A) / n;
nv = 0;
for (i = 0; i < n; i++) {
double x = A + i * (B - A) / n;
double y = a*x*x + b*x + c;
nv += dx * y;
}
} while(fabs(nv - sv) > 0.0005 * fabs(nv + sv));
printf("Surface: %lf\n", nv);
return 0;
}
The code is well-behaved for empty intervals (where A = B) or reversed intervals (where A > B). The inpt is still quick and dirty. It should really heck that the entered values are valid numbers. There's no need to restrict the input arbitrarily, though.