Assume you want to make sure that the user enters a positive number that is divisible by 10 with no remainder. Write the condition you would use in the following do-while loop.
do {
cout << “Enter a positive number that is divisible by 10 with no remainder” << endl;
cin >> number;
}
while ( ____________________________________________________________);
Does this program solve the problem of testing an integer to be divisible by 6?
Code:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Chapter7Problem5
[Code] ....
Create a program that keeps on reading positive numbers until user enters a zero value, and then outputs the last positive number entered.
I do not know how to output the 'last positive number' and I do not even know am I doing things correctly so far....
Code:
#include <stdio.h>
int main() {
double num;
for(;;) {
printf("Enter a positive number: ");
scanf("%lf",&num);
[Code] ....
I was going through the following statement in a c book:
"interpreting -1 as unsigned int leads to a big +ve number"
-1 is already unsigned .... then what does the above statement mean ??
My program uses a while loop to eventually get to an error of zero and a root of sqrt(3). I'm not understand why after the third iteration the program fails to compute a new x value. I'm using Visual Studio 2013. The code tag instructions were dubious.
Code:
#include <stdio.h>
#include <math.h>
main() {
/*This program uses the Newton-Raphson method to solve y = (x^3)-3 for it's roots.*/
printf("This program uses the Newton-Raphson method to solve y = (x^3)-3 for it's roots. Enter your estimate of the root.
");
float x,y,z;
int num;
num = 0;
[Code]...
#include <iostream>
using namespace std;
int main() {
int h;
double A[10][10];
[Code] .....
Write a C++ program that will input from the user a positive number n and find its factorial. Don’t forget to validate the input. The factorial of a positive integer n (denoted by n!) is defines as the product of the integers from 1 to n.
n! = 1* 2 * 3 * ... * (n - 1) * n
You should allow the user to continue working with your program for additional data sets.
Sample output:
Please enter a number: 5
5! = 120
Would you like to continue (Y/N)?Y
Please enter a number: 3
3! = 6
Would you like to continue (Y/N)?N
Good Bye!!
My code for what i think I'm doing is as follows:
#include <iostream>
using namespace std;
int main(){
int i=1;
int n;
[Code] ....
So I tried creating a test code for reading a negative number and positive number but whenever I enter a negative number it read it as being positive.
#include <stdio.h>
#include <iostream>
#include <iomanip>
[Code].....
PS: I am using char over int because the program that I am testing for requires me to use 8 bit variable.
Code:
#include <stdio.h>
int main(){
int A, B;
char decision;
printf("Do you have an integer to input? [Y/N]: ");
scanf("%c",&decision);
if(decision=='Y' || decision=='y'){
[Code]....
After entering a single integer, it doesn't scan again for another integer. What's wrong?
I'm using a mac btw, if that makes a difference with Ubuntu/Linux.
Write a program that displays all the numbers from 100 to 1,000, ten per line, that are divisible by 5 and 6. Numbers are separated by exactly one space.
#include <iostream>
#include <stdlib.h>
using namespace std;
[Code]....
How to write a program of odd numbers divisible by n
View 1 Replies View Related#include <iostream>
using namespace std;
int main() {
int i=0;
while (i<=100) {
[Code] .....
the part where i need to check for values divisible by 5 is incomplete.
I want to find the remainder of the division between a and b, but without using the reminder operator a%b.I thought to subtract b from a as long as a>b, that will give the remainder, but I don't know how to write it in code.
View 11 Replies View RelatedI've pretty much finished the entire program, except for the actual calculation part.
"Given a range of values determine how many integers within that range, including the end points, are multiples of a third value entered by the user. The user should be permitted to enter as many of these third values as desired and your output will be the sum of total multiples found."
I've defined functions to take user input for the low range, high range and a do-while loop to take as many third inputs as the user wants (terminated by entering -1, as requested by the question)
To actually calculate if they're divisible, I found out that if A%B = 0, then they are divisible, so I thought I would create a loop where each value in the range between A and B is checked against the third value to see if they output a zero.
What I need to end up with is a program that tells the user how many integers are divisible by the numbers in the range, i.e: "Enter the low range value: 335 Enter the high range value: 475 Enter a value to check within the range: 17 Enter a value to check within the range: -1 There are 8 total values that are divisible by the numbers in the range." Going back to my original question, how would I create a loop or something to "check" how many values are equal to zero, and consequently increment a variable for each instance? (This is how I think it should be done)
Code:
#include <stdio.h>
//GLOBAL DECLARATIONS
int getlowR();
int gethighR(int);
[Code].....
A program that outputs all numbers divisible by both 5 and 6.This is what i have written :
#include<stdio.h>
int main () {
int i=200;
int b=10;
while(i<1000)
i b/6%==0
}
where could i have gone wrong.
Display the remainder of the square of numbers from 100 to 10. This square of numbers must be divisible by the numbers from 100 to 10 respectively. what i need to in this
View 3 Replies View RelatedI have to build a program that calculates the remainder of the expression
(2^10)!/((2^10-500)! * 500!)
when divided by 10^9+7
where x^y = x*x*x*x...*x (y times)
and x! = x*(x-1)...*1
How can I do that? I know how to calculate the remainder of x! and the remainder of y!, but I do not know how t calculate the remainder of x!/y!. I can´t even store this in a variable because x! is very large.
if i have two integers, say number1 and number2, stored in arrays where each index is a digit of the number (i.e. if my numbers are 321 and 158, then number1 = {3,2,1} and number2 = {1,5,8}), can i find the remainder of number1/number2? assume number1 > number2.
View 1 Replies View Relatedwhere are operators defined in C/C++? in headers or compiled code?
looking for the definition of how % modulus is calulated
So I am using Visual Studio 2012 Professional, this is C++ code. I am just trying to get the remainder from a very simple division. Nothing difficult, heres the code:
double getProbability(){
int rd = random();
int max = numeric_limits<int>::max();
double result = rd % max;
cout << "Probability: " << result << "
";
return result;
}
When I look at the values in debug I get:
max 2147483647
rd 1804289383
result 1804289383.0000000
That is completely wrong. The answer should be 0.840188. What is going on here?
random() just returns a number from a vector that was prepopulated with "random" integers. Not really random, but that isn't all that important. What is important is why on earth is a % operation returning such a huge number. I assigned the values to variables so I could look at them in the debugger. I know I am going to probably get a thousand different ways that I could do this "better" but again, that isn't what I am looking for. I would just like to know why the % operation is doing what it is doing?
For class I need to write a program that inputs a file (the dividend), performs binary division on the file (using 0x12 as the divisor), and outputs the remainder(checksum).
I have researched binary division algorithms and I get the general gist, but I'm still unsure where to start. How would I store the dividend and divisor? As two arrays of bits?
Then, I need to figure out how to perform shifts and XORs on the the binary numbers. Maybe I should use bitwise operations?
Say I wanted to overload the modulus operator to return the remainder of a division between two floating point numbers. Why isn't a custom double operator%(double, double) allowed even though that function isn't available in the standard anyway?
View 5 Replies View RelatedWhat are positive and negative infinity for different data types in c++, are they represent maximum and minimum limit of a type? or positive infinity is not a finite value.can some explain this positive and negative infinity paradigm
View 3 Replies View RelatedI have written a function that takes in a positive decimal and returns its Binary equivalent; however, the output always adds an additional zero to the binary. What could I do to get rid of it?
If the number is 7, it outputs 0111 instead of 111.
Code:
#include <stdio.h>
void Dec(int n) {
if(n > 0)
Dec(n/2);
printf("%i", n%2);
[Code] ....
I have task to make program.. To enter 10 positive numbers in an array. Find three numbers that stand near each other and that have the maximum sum.
I understand how to make array, enter 10 number i find only one biggest number from 10.. How to find three max number sum that stand near each other..
int Fib1 = 1;
int Fib2 = 2;
int Fib3 = 0;
int randomynumber;
int Loop;
[code].....
this returns negative numbers sometimes.what did i do wrong side note this is not the complete program it is only the part with the problem because the complete code is sort of longish and very confusing