I need to create a program that displays the sum of the even integers between and including two numbers entered by the user.
This is what I have so far and it's not working!?
So point me in the right direction please!
//Advanced30.cpp - displays the sum of the even integers between and
//including two numbers entered by the user
//Created/revised by <your name> on <current date>
#
include <iostream>
using namespace std;
int main()
{
// declare variables
int num1 = 0;
int num2 = 0;
int sum= 0;
cout << "Enter the First Number:" << endl;
cin >> num1;
cout << "Enter the Second Number:" << endl;
cin >> num2;
for (num2 = num1; num1 <= num2; num1 += 2) sum += num1;
num1 = num1 % 2 == 0 ? num1 : num1+1;
num2 = num2 % 2 == 0 ? num2 : num2-1;
return 0;
Try to do EXACTLY what your computer is doing when it's doing the loop. Do it on a paper. Keep track of num2, num1 and their value.
You'll see very quickly where the problem is.
try the Loop
for(; num1<=num2;num1++)
{
if(num1%2==0)
sum=sum+num1
}
for (num2 = num1; num1 <= num2; num1 += 2) sum += num1;
You've overwritten your stop-point. :)
I would also suggest more meaningful names:
int start=0;
int real_start=0;
int stop=0;
int sum=0;
/* ... */
real_start = (start % 2) ? start+1 : start;
for (int i = real_start; i <= stop; i+=2) sum += i;
/* ... */
Related
I'm making a program where a person can input a direction and in the if statement, it adds/subtracts x/y axis and it draws a line after it gets over. The problem is that for some reason, it only works at the first iteration and doesn't draw any more lines after that.
I added a cin >> x >> y to test it out but it only draws one line and doesn't draw anymore.
Initially, the choices were in a switch statement but I changed to if because I thought that was causing the error.
#include "pch.h"
#include <iostream>
#include <fstream>
#include <string.h>
#include <cairo.h>
#define _USE_MATH_DEFINES
#include <math.h>
using namespace std;
char b = NULL;
char u = 'œ';
char d = 'd';
int main()
{
cairo_surface_t *surface = cairo_image_surface_create_from_png("background.png");
cairo_t *cr = cairo_create(surface);
cairo_set_source_rgb(cr, 0, 0, 0);
cairo_set_line_width(cr, 5);
double x = 0, y = 240;
cairo_move_to(cr, x, y);
long int count = 0;
int cl = 0;
int crr = 0;
int choice = 0;
int n;
system("cls");
while (choice != 5)
{
cin >> x >> y;
cairo_line_to(cr, x, y);
cairo_stroke(cr);
cairo_surface_write_to_png(surface, "spiral.png");
cout << "Current no. of points are : " << count << "/4096" << endl;
cout << "Enter direction: \n" << endl;
cout << "1 - Top Left \t 2 - Up \t 3 - Top Right " << endl;
cout << "4 - Left \t 5 - Stop \t 6 - Right" << endl;
cout << "7 - Bot. Left \t 8 - Down \t 9 - Bot. Right" << endl << endl;
cout << "Enter you choice: ";
cin >> choice;
if (choice == 1)
cout << "Test";
else
{
//More choices include the direction the person needs to go and it subtracts/adds to the x/y part
cout << "How many times ?: ";
cin >> n;
for (int i = 1; i <= n; i++)
{
x++;
count++;
cl++;
if (cl == 256)
{
cl = 0;
crr++;
}
}
system("cls");
}
}
}
I expect it to draw lines to a particular direction. Say the person inputs right, it draws a line towards right and so on. But here, no lines get drawn at all (except if I add a cin >> x >> y at the start of the while loop, that draws one line and that's it, no more lines.)
This fails because there is no current point anymore. After cairo_stroke(cr);, you can add cairo_move_to(cr, x, y); and it should start drawing more lines in the way you expect. I think... I'm not quite sure what you are up to with this program.
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How would you be able to replace digits in a given number using basic c++. Example if the number is 23444 and you want to take the old digit 4 and replace is with a new digit 5 to get a new number 23555.
I have some work done below but when I enter the inputs, it ends up giving me an incorrect result.
cout << "Enter the number: " << endl;
cin >> number;
cout << "Enter the old digit: " << endl;
cin >> oldDigit;
cout << "Enter the newDigit: " << endl;
cin >> newDigit;
newDigit=oldDigit;
cout << Newnum << endl;
You can convert int to char* using itoa() and iterate over it to
check does it contain number. If it does, get 4's position and replace
it with 5.
I know you didnt work with strings, but it can be helpful in your case.
Simple code:
#include <string.h>
#include <iostream>
#include <stdlib.h>
int main(){
int numer;
std::cin>>numer;
char* str;
itoa(numer, str, 10);
for(int i = 0; i < strlen(str); i++){
if(str[i] == '4') str[i]='5';
}
}
If I understand you correctly, you don't just want to simply add 111, you want to treat the number as a string, then change elements in the array. Is that correct?
This may get you on the right track:
Convert an int to ASCII character
if you really want to use only int to do this, here is a working example (base on some of your code)
#include <iostream>
using namespace std;
int replaceDig( int num, int oldDigit, int newDigit)
{
if(num==0)return 0;
int digit = num%10;
if(digit==oldDigit)digit = newDigit;
return replaceDig(num/10,oldDigit,newDigit)*10+digit;
}
int main()
{
int num, newnum, oldDigit, newDigit;
cout << "Enter the number: " << endl;
cin >> num;
cout << "Enter the old digit: " << endl;
cin >> oldDigit;
cout << "Enter the newDigit: " << endl;
cin >> newDigit;
newnum = replaceDig(num, oldDigit, newDigit);
cout << newnum << endl;
return newnum; //do you really want to return this?
}
I have come up with a solution . Don't know if it contains bug or not. Please let me know.
int num = 23444 ,new_num = 0;
int mod;
int exponent = 0;
/**
* Now form the new number
*/
while ( num > 0 ) {
mod = num % 10;
num /= 10;
if ( mod == 4 ) // check whether this is the old digit or not
new_num += 5 * pow( 10 , exp); // replace with new digit
else
new_num += mod * pow(10 , exp); // otherwise no change
exp++;
}
num = new_num;
std::cout << num;
I hope this works for you -
std::string s = std::to_string(23444);
std::replace( s.begin(), s.end(), '4', '5');
int num = std::stoi(s);
int replaceDig( int num, int oldDigit, int newDigit) // replacing the old digits in a number with a new digit
{
int position = numDigits(num);
int remainder = num;
int currentDigit;
while (remainder >0)
{
currentDigit=(num/pow(10,position))%10;
if(currentDigit==oldDigit)
{
num = num - oldDigit*pow(10,position);
num = num + newDigit*pow(10,position);
}
remainder = remainder/10;
position--;
}
}
This is the general idea, I guess. I didn't try to compile it though. And of course, this version isn't really optimized and we could find some more efficient ways of doing it. Oh, and it doesn't work with negative numbers, but this should be quite easy to adapt.
I can't see my output from this code I have written. I'm trying to calculate the mean and standard deviation of a set of numbers from a file. I'm lost as to what the problem is and I won't know if my code is right until I can see output. Here's what I have so far:
#include <iostream>
#include <cmath>
#include <fstream>
using namespace std;
int main()
{
// Declare Variables
int n;
int xi;
int sdv;
int sum;
int sum2;
int sum3;
int mean;
// Declare and open input files
ifstream inData;
inData.open("score.dat");
if (!inData) // Incorrect Files
{
cout << "Cannot open input file." << endl;
return 1;
}
// Initialize variables and output
inData >> xi;
n = 0;
sum = 0;
sum3 = 0;
sdv = 0;
mean = 0;
while (inData)
{
sum += xi;
sum2 = sum * sum;
sum3 += (xi * xi);
mean = sum / n;
sdv = (sum3 - sum2) / (n * (n - 1));
inData >> xi;
}
// Print commands
cout << "The Standard Deviation of the Tests is:" << sdv << endl;
cout << "The Mean of the Tests is: " << mean << endl;
inData.close();
system("pause");
return 0;
}
After running your code I found a few bugs. You probably don't get any output because the program crashes the first time through the while loop on the line:
mean = sum / n;
due to a divide by zero error.
The other bug is that you don't increase (increment) the value of n in your loop, so you always only have one number.
Adding an n++ at the beginning of your loop will fix that
while (inData)
{
n++;
sum += xi;
...
But you still get a divide by zero on the sdv when n == 1:
sdv = (sum3 - sum2) / (1 * (1 - 1));
if you add a condition before the division it will work:
if (n >= 2)
sdv = (sum3 - sum2) / (n * (n - 1));
Look into debugging tools like gdb to help catch things like this.
Using Bottom to up DP approach, I am able to solve the problem How to solve http://www.spoj.com/problems/MST1/ upto 10^8.
If input is very large n upto 10^9. I will not be able to create lookup table for upto 10^9. So what will be better approach to solve the problem ?
Is there any heuristic solution ?
#include <iostream>
#include <climits>
#include <algorithm>
using namespace std;
int main()
{
const int N_MAX = 20000001;
int *DP = new int[N_MAX];
DP[1] = 0;
for (int i = 2; i < N_MAX; i++) {
int minimum = DP[i - 1];
if (i % 3 == 0) minimum = min(minimum, DP[i/3]);
if (i % 2 == 0) minimum = min(minimum, DP[i/2]);
DP[i] = minimum + 1;
}
int T, N; cin >> T;
int c = 1;
while (T--) {
cin >> N;
cout << "Case " << c++ << ": " << DP[N] << endl;
}
delete[] DP;
}
#include <iostream>
using namespace std;
int main ()
{
int size = 0;
int* myArray = new int [size + 1];
cout << "Enter the exponent of the first term: ";
cin >> size;
cout << endl;
for (int i = size; i >= 0; --i)
{
cout << "Enter the coefficient of the term with exponent "
<< i << ": ";
cin >> myArray[i];
}
for (int i = size; i >= 0; --i)
{
cout << i << endl;
}
return 0;
}
Why am I getting an assertion error on input greater than 2? This is the precursor to a polynomial program where the subscript of the array is the power of each term and the element at array[subscript] is the coefficient.
Your array is allocated to be an int[1]. It needs to be allocated after you read in the size value.
You are initializing your array when size = 0, giving an array size of 1
You get your assertion error when you go outside of the array bounds (1).
myArray always has size 0 + 1 = 1. i starts out at whatever the user inputted, and the first array access you make is myArray[i]. So, say the user inputs 5, your array has size 1 and you access myArray[5]. It will fail!
I would allocate the array AFTER you input size.