Want to print for example
Grade: 4/5
80 percent
Program ask a user how many math problems they want to solve and prints out the number of "wrongs/the number of rights" and their grade. I think I dont have my math right at the end of the code cause its printing out for example:
Grade: 4/5
-7446528 percent
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
# include <stdio.h>
int main()
{
int NumberOfTimes, AddAns, SubAns, AddCorrect=0, SubCorrect=0, CorrectAnsAdd, CorrectAnsSub, TotalCorrect, TotalWrong, Add;
int i,a,b,c,d,e,f,g;
float percent;
printf("\n");
printf("-------------------MATH QUIZ------------------------\n");
printf("Enter the number of Math problems you want to solve:"); //enters the # of problems the program produces
scanf("%d", &NumberOfTimes);
printf("\n");
srand(time(NULL));
//Random number generator
for (i=0;i<NumberOfTimes;++i)
{
b = rand() %3 + 1;
c = rand() %3 + 1;
a = rand() %2 + 1;
//Random addition problems
if (a == 1)
{
printf("%d + %d = ", b,c);
scanf("%d", &AddAns);
d = b + c;
if (AddAns == d)
{
printf(" +Correct\n");
AddCorrect = AddCorrect + 1;
}
//Random subtraction problems
else
{
printf(" +Wrong, it was %d\n", d);
AddIncorrect = AddIncorrect + 1;
}
}
if (a == 2)
{
printf("%d - %d = ", b,c);
scanf("%d", &SubAns);
g = b - c;
//Produces right or wrong answers
if (SubAns == g)
{
printf(" +Correct\n");
SubCorrect = SubCorrect + 1;
}
else
{
printf(" +Wrong, it was %d\n", g);
SubIncorrect = SubIncorrect + 1;
}
}
}
//Producing the output to wrong/right numbers and grade percentage
TotalCorrect = AddCorrect + SubCorrect;
printf("\n");
printf("Grade: %d/%d\n",TotalCorrect,NumberOfTimes);
printf("\n");
percent=NumberOfTimes/TotalCorrect;
printf("%d percent \n", percent);
return 0;
}
There are a few things going on. First, to print a float with printf, use %f. %d is for ints.
Secondly, when you calculate percent, you're accidentally using integer division. Since NumberOfTimes and TotalCorrect are both integers, NumberOfTimes/TotalCorrect performs integer division and produces an int. It's only converted to a float after the whole initializing expression is evaluated. Use this instead:
percent = (float)TotalCorrect / NumberOfTimes;
// OR, if you want an actual percent:
percent = 100.0f*TotalCorrect/NumberOfTimes;
Then, using %f:
printf("%f percent\n", percent); // "80.000000 percent"
Note that this will display the percentage out to many decimal places; if you want a cleaner display without a decimal point, you could just calculate the percent as an int:
// multiply before dividing to avoid integer division problems
int percent = 100*TotalCorrect/NumberOfTimes;
printf("%d percent\n", percent); // "80 percent"
Hope this helps!
In C, dividing two integers won't get you a floating point number, even if the assignment is to a float. It's just how the language works. You'll have to cast NumberOfTimes and TotalCorrect as floats.
So replace
percent=NumberOfTimes/TotalCorrect; with
percent=(float)TotalCorrect/(float)NumberOfTimes * 100;
Furthermore, you're trying to print a float as an integer in the line
printf("%d percent \n", percent);
which is giving you a wonky result. Instead, try:
printf("%d percent \n", (int)percent);
Related
i have a homework but i cant get the answer
I need to write a program in C...
Here is what is needed: You need to enter "n" natural number as input , and from all the natural numbers smaller than "n" , its needed to print the number which has the highest sum of devisors.
For exp: INPUT 10 , OUTPUT 8
Can anyone help me somehow?
I would really appreciate it !
i tried writing a program for finding the devisor of a number but i cant get far from here
#include <stdio.h>
int main() {
int x, i;
printf("\nInput an integer: ");
scanf("%d", &x);
printf("All the divisor of %d are: ", x);
for(i = 1; i < x; i++) {
if((x%i) == 0){
printf("\n%d", i);
}
}
}
I have implemented using function which will takes input number from user and then return the sum of divisor. hope this is one you looking for
/* function to return of sum of divisor
** input: x: integer number from user input
** return sum: sum of divisor of x
*/
int sum_of_divisor(int x)
{
int sum = 0;
for(int i = 1; i < x; i++)
{
if((x%i) == 0)
{
printf("%d\n", i);
sum = sum+i;
}
}
return sum;
}
int main() {
int x, i;
printf("\nInput an integer: ");
scanf("%d", &x);
printf("All the divisor of %d are: ", x);
printf("the sum of divisor is %d ", sum_of_divisor(x));
return 0;
}
Output:
Input an integer: 10
All the divisor of 10 are: 1
2
5
the sum of divisor is 8
After checking if i is a divisor of x, you should then store that value in another variable, for example m.
Repeat until a new divisor i is higher than that number. Add this new value to m.
//Enter a 4-digit integer n from the keyboard, and write a program to divide it into two 2-digit integers a and B. Calculate and output the results of the addition, subtraction, multiplication, division and redundancy operations of the split two numbers. For example, n=-4321, if the two integers after splitting are a and b, then a=-43 and b=-21. The result of division operation requires that it be precise to 2 decimal places, and the data type is float. Redundancy and division operations need to take into account the division of 0, that is, if the split B = 0, then output the prompt information "The second operator is zero!"
//Failure to pass the test，how should i fix
#include<stdio.h>
#include<math.h>
int main()
{
int x, a, b;
printf("Please input n:\n");
scanf("%d", &x);
a = x / 100;
b = x % 100;
printf("%d,%d\n", a, b);
printf("sum=%d,sub=%d,multi=%d\n", a + b, a - b, a*b);
if (b == 0)
printf("The second operater is zero!");
else
printf("dev=%.2f,mod=%d\n", (float)a / b, a%b);
}
You forgot to check that x is a 4-digits number. So if the input is 12345 or 123 you don't satisfy the requirement.
#include <stdio.h>
int main()
{
int x, a, b;
int passed = 0;
// Enter a 4 digits number: ABCD
do {
printf("Enter X = ");
scanf("%d", &x);
passed = (x >= 1000 && x <= 9999) || (x >= -9999 && x <= -1000);
} while (!passed);
a = x / 100;
b = x % 100;
printf("Numbers: %d %d \n", a, b);
printf("Sum = %d \n", a + b);
printf("Sub = %d \n", a - b);
printf("Mul = %d \n", a * b);
if (0 == b) {
printf("Div by Zero \n");
} else {
printf("Div = %f \n", (double)a / b);
printf("Mod = %d \n", a % b);
}
return 0;
}
We have been learning about recursion vs iteration in C this week and we were required to make a program that recursively determines the value of the nth term of a geometric sequence defined by the terms a, ar, ar^2, ... ar^(n-q).
For the most part, I think I have it figured out, as it seems to display the correct values per run, but it doesn't manage to break the recursion when the tested value reaches zero. Also, if possible to get a better explanation of recursion, and some examples of when recursion would be preferred over iteration as I'm still struggling with the concept.
// 2/20/2018
//Lab 6 Solution for Page 369 PE 4 B
//including libraries to be used
#include <stdio.h>
#include <math.h>
int main() {
//Function prototype
double goAnswer(int *, double, double, double, double *, int);
//Declaring variables
int nValue = 0;
double ratio = 0;
double firstTerm = 0;
double answer = 0;
double addedAnswer = 0;
int count = 1;
//Setting up to ask for each value
printf("Please enter in the value of n: ");
scanf("%d", &nValue);
printf("Please enter in the ratio you'd like to use: ");
scanf("%lf", &ratio);
printf("Please enter in the first term to use: ");
scanf("%lf", &firstTerm);
addedAnswer = goAnswer(&nValue, ratio, firstTerm, answer, &addedAnswer,
count);
//Printing out the value of the first nth terms
printf("The value of all terms added together is: %lf\n", addedAnswer);
return 0;
}
//function header
double goAnswer(int *nValue, double ratio, double firstTerm, double answer,
double *addedAnswer, int count) {
if (nValue == 0){
return 0;
}
else{ //This part calculates the answer, prints the value to the screen,
adds the answer to a running sum, decreases the nValue by one and calls the
function again with the lower nValue
answer = firstTerm * pow(ratio, count);
printf("The value of term %d is: %lf\n", count, answer);
printf("This is the nValue: %d \n", *nValue);
*addedAnswer += answer;
nValue -= 1;
return (goAnswer(nValue, ratio, firstTerm, answer, addedAnswer,
(count + 1)));
}
}
I've just started to learn coding
#include <stdio.h>
int main(void)
{
int i=0, x=0;
for (i=1; i<=100; i++) {
x++;
if (x%5==0 || x%3==0)
printf("The numbers are : %d\n", &x);
}
return 0;
}
so I'm trying to print all the integers <=100 that are divisible by either 3 or 5.
In the line:
printf("the numbers are : %d", &x);
printf is expecting an integer because of the %d format specifier, you have given it the address of an integer, thats what the & means, address of x. To fix this give printf what it desires, an int:
printf("the numbers are : %d", x);
The ampersand (&) operator returns the address of a variable, which isn't what you want - you just want the value. Also, note that you don't need two variables (x and i) - you can just use the loop's counter:
printf("The numbers are:\n");
for (i = 1; i <= 100; i++) {
if (i % 5 == 0 || i % 3 == 0) {
printf("%d ", i);
}
}
Removing the & will eliminate the error. You may also want to move the "the numbers are:" statement outisde of the loop so the text does not print each time through the loop.
I have a problem, I tried to write a program to show the whole sum from 1 to 22 and after that, to do 2 while loops. The first one is supposed to perform the sum of some numbers given by the user, as an example: you type 10, 30 and 40 then as you enter a 0 the program sums the first three numbers. Unfortunetly the first while loop is not working. It goes directly to the last while loop where it is supposed to type a decimal numbers like (10.20 30.50 40.55) and after you type 0 again it sum those numbers and add and multipli every entry with 1.19. So far the last loop is working properly, unfortunately the second loop does not, if I move printf and scanf over the while it let me write but just start writing w/o stopping the number I wrote . Thank You in advance!
Here is the code :
#include <stdio.h>
int main()
{
int sum = 0;
int a;
int b;
double i;
double sum1 = 0;
for (a= 0; a <= 22; a++) {
sum = sum + a;
printf("the sum from 1 till 22 : %i\n ", sum);
}
while (b != 0) {
printf("type a number:");
scanf("%i", &b);
sum += b;
printf("%i\n", b);
}
printf("the sum is : %i\n", sum);
while(i !=0) {
printf ("Type a decimal number:");
scanf ("%lf",&i);
sum1 += i*1.19;
printf("%lf\n", i);
}
printf("The decimal summ is: %lf\n",sum1);
return 0;
}
You don't initialise i to any value before entering the loop with
while(i != 0)
i might very well be zero at this point, so your loop won't be entered even once. Initialising i to a non-zero value should fix this particular problem. The same holds for the variable b.
You should turn on warnings in your compiler, so it can show you problems like this one.
The first time the condition of the second while is evaluated, b has undefined value, since it wasn't initialized. The same applies to the third while.
Whether or not both loops are executed is only a question of chance.
Initialize both variables with non-zero values to ensure both whiles are entering. Or use a do-while:
do {
printf("type a number:");
scanf("%i", &b);
sum += b;
printf("%i\n", b);
} while (b != 0);
Don't test b with while, test it after the user enters the number. Then you can use break to exit the loop.
while (1) {
printf("type a number:");
scanf("%i", &b);
if (b == 0) {
break;
}
sum += b;
printf("%i\n", b);
}
while(1) {
printf ("Type a decimal number:");
scanf ("%lf",&i);
if (i == 0.0) {
break;
}
sum1 += i*1.19;
printf("%lf\n", i);
}
Your only issues are initialization: see edits in the code below. (it compiles and runs)
Did you get any compiler warnings for these? If not, you should change your settings so you do.
#include <stdio.h>
int main()
{
int sum = 0;
int a;
int b=-1; //initialize (any non-zero value will work)
double i;
double sum1 = 0;
for (a= 0; a <= 22; a++) {//a initialized in for(...) statement, (this is good)
sum = sum + a;
printf("the sum from 1 till 22 : %i\n ", sum);
}
while (b != 0) { //b Needs to be initialized before using (done above)
printf("type a number:");
scanf("%i", &b);
sum += b;
printf("%i\n", b);
}
printf("the sum is : %i\n", sum);
i=-1; //initialize i to any non-zero value
while(i !=0) {
printf ("Type a decimal number:");
scanf ("%lf",&i);
sum1 += i*1.19;
printf("%lf\n", i);
}
printf("The decimal summ is: %lf\n",sum1);
getchar();
return 0;
}