I am trying to solve ax + by = n.
When I put n = 7, it solves the equation correctly as X = 2 and Y = 1.
But when I put n = 1, it does not solve the equation. Even though, this equation has valid integer solution, X = 17, Y = -11. here is my full program.
#include <iostream>
using namespace std;
void PrintXY(int a, int b, int n)
{
for (int i = 0; i * a <= n; i++) {
if ((n - (i * a)) % b == 0) {
cout << "x = " << i << ", y = "
<< (n - (i * a)) / b;
return;
}
}
cout << "No solution";
}
int main()
{
int a = 2, b = 3, n = 1;
PrintXY(a, b, n);
return 0;
}
Output when n = 7:
x = 2, y = 1
Output when n = 1:
No solution
Reasoning.
2*(2) + 3*(1) - 7 = 4 + 3 - 7 = 0
2*(17) + 3*(-11) - 1 = 34 - 33 - 1 = 0
Both equations solve to give 0. But what is wrong in my program that is causing it to give "No Solution".
The problem is with the termination condition:
i*a<=n
This(n>=a*i) need not be true, and is especially not true in case of the solution (ie X=17, Y=-11). Which seems reasonable - Without any bounds(limits) on the answer(either X or Y) , how would you find the solution to a linear equation(with an infinite possible range) in a closed for loop ?
I have made a code to solve a problem.
This is the link of problem: https://brilliant.org/practice/bijections-overview/?p=2
My opinion is use recursion. I have 3 elements to sum up 5, I iterate single element from 0 to 5 and so on.
This is my code:
#include<bits/stdc++.h>
using namespace std;
int count_sum_5(int sum, int n)
{
if(n == 1) return 1;
for(int i = 0; i <= sum ; i++){
return 1 + count_sum_5(sum - i, n - 1);
}
}
int main()
{
int sum = 5;
int count_ele = 3;
cout << count_sum_5(sum, count_ele);
}
Its output is 3, I think it only runs on i = 0 but does not run on i = 1, 2, 3, 4, 5. Could you help me?
Your function will return 1 if n is 1. Otherwise it will enter the for cycle, which will return the value of the first iteration. You call your function by passing 5 and 3, respectively.
First call: It returns 1 + count_sum_5(5, 2)
Second call: It returns 1 + count_sum_5(5, 1)
Third call: It returns 1
So, the second call will evaluate to 1 + 1 = 2 and then the second call will evaluate to 1 + 2 = 3. I'm not sure what your intention is, but the for cycle is not needed for your recursion. Your function is equivalent to
int count_sum_5(int sum, int n)
{
if(n == 1) return 1;
return 1 + count_sum_5(sum, n - 1);
}
I'm not sure whether you are satisfied with this. If not, then please let me know what you intend to achieve, sample inputs and outputs would help. If you provide further information, then I will edit my answer accordingly.
return evaluates 1 + count_sum_5(sum - i, n - 1) and then immediately exits the whole function, on the first iteration of the loop. It's not possible to return from a function multiple times.
I don't know what exactly are you trying to solve using that. But For loop will run only one time for i=0, because as soon as it evaluates the 1+count_sum(sum-i, n-1), It will return it.
However, you can do the following changes:
#include<bits/stdc++.h>
using namespace std;
int count_sum_5(int sum, int n)
{
if(n == 1) return 1;
int temp = 0;
for(int i = 0; i <= sum ; i++){
temp += (1 + count_sum_5(sum - i, n - 1));
}
return temp;
}
int main()
{
int sum = 5
int count_ele = 3;
cout << count_sum_5(sum, count_ele);
}
I was writing a small snippet to get a Fibonacci number sequence depending on the user input. If the user supplies 4 as an input, it should return him the first N members of the Fibonacci sequence.
#include <iostream>
using namespace std;
int main (){
int a = 0;
int b = 1;
int c;
int n = 3;
n -= 2;
if (n == 1){
cout << a << endl;
} else {
cout << a << b << endl;
for (int i=0;i<n;i++){
c = b + a;
cout << c << endl;
a = b;
b = c;
}
}
}
However, I end up getting a 0 as an output for whatever number I supply. I have this working in PHP and I kinda miss where I've blundered. I guess I don't actually render input and output properly.
int a =0;
int n = 3;
n -= 2;
if (n == 1){
cout << a << endl;
}
You have n equal to 3, you subtract 2, thus n equal to 1, so, you enter the if body and output a, which is zero.
[EDIT]
You don't seem to get any input -as stated in a comment- in your program (you could use std::cin or std::getline() for this), but you probably mean that you have the input hard-coded, by changing the value of n by hand.
You may want to check how the Fibonacci series program is expected to work:
Fib. at Rosseta page.
Fib. with recursion
Non-recursive Fib.
After reading the links I provided above, you should be able to see that your code should be changed to this:
#include <iostream>
using namespace std;
int main (){
int a = 1;
int b = 0;
int c;
int n = 10; // "input" is 10
if (n == 0 || n == 1) { // 0 and 1 case
cout << n << endl;
} else {
for (int i = 2; i <= n; ++i) { // here you want to reach n
c = a + b;
b = a;
a = c;
}
cout << c << endl;
}
return 0;
}
However, the code above outputs only the result. You should slightly modify it to get the terms of the sequence, but I'll leave you have some fun too.
In order to really let the user input the number, change:
int n = 10;
to
int n;
std::cout << "Please, input.\n";
std::cin >> n;
However, letting user inputting must be followed by validation of the input. You see users can, by accident or not, provide input in your program, that can cause undefined behaviour.
The sequence you want is 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, ...
As I pointed out in a comment to another answer, your code does not produce a correct Fibonacci sequence. F(3) isn't the problem with your code; the problem is that you get confused between all the variables, a, b, c and use them to mean different things at once.
You also incorrectly decrement n: your code does it in the wrong place, and even if you move it to the right place, it wouldn't help as the operation would make n go negative.
Your existing Code
Let's walk through your code a bit:
int a = 0;
int b = 1;
int c;
int n = 3;
n -= 2;
Well, this is weird. We set n to 3 then immediately subtract 2, making it 1. This means that if you try to set n to 0, 1, or 2 you end up with n being a negative number. If you set it to 3, you end up with n being 1.
if (n == 1){
cout << a << endl;
}
We're in trouble right here. Remember that you subtract 2 from n which means that for n==3 you will return whatever is in a which is wrong. But even if you meant this to special-case F(1) that code is still wrong because F(1)=1.
else {
cout << a << b << endl;
for (int i=0;i<n;i++){
Remember, that we can get here with n zero or negative. Obviously in the case of n <= 0 this loop will never execute, so c will never be printed.
c = b + a;
cout << c << endl;
Here, we seem to calculate and output the next Fibonacci number by adding the two previous numbers. This should be fine.
a = b;
b = c;
And here, we keep the new Fibonacci number and its predecessor for the next loop iteration, if any.
The problems with this code are, of course, fixable. But the problem is that the existing code is confusing. It outputs all sorts of different values, and it's unclear what variable is supposed to represent.
Looking at this problem, your first instinct would be to make a function which accepts as input a number n and returns F(n) - you could call it fib or somesuch.
Reworking this
So, how to go about writing such a function? Here's a simple recursive implementation that you can use:
int fib(int n)
{
if ((n == 0) || (n == 1))
return n;
return fib(n-1) + fib(n-2);
}
Notice how this function is short, sweet and to the point. There's no need for a ton of variables, no need for complicated control structures or storing state. It almost reads like a text-based description of the Fibonacci algorithm.
Of course, it's not super-efficient and ends up redoing a lot of work. That's a legitimate criticism, but it's unlikely that there performance considerations here.
Still, perhaps you just don't like recursion. Many people think of recursion as a dirty word, and avoid it with a passion. So how about a non-recursive implementation instead? It's possible, but it's a bit more difficult to understand.
int fib (int n)
{
/* F(0) = 0 */
if (n == 0)
return 0;
int a = 0;
int b = 1;
for (int i = 2; i < n; i++)
{
int c = a + b;
a = b;
b = c;
}
/* F(n) = F(n-2) + F(n-1) */
return a + b;
}
This is a little bit more efficient and not that much more difficult to understand.
I hope that this helped.
Try this which would give you the list you needed.
#include <iostream>
using namespace std;
int fib(int num){
int ans;
if (num >2) {
ans = fib(num-1) + fib(num-2);
}
else
ans = 1;
return ans;
}
int main()
{
int num, x=1;
cin >> num;
while (num >= x) {
cout << fib(x) <<" ";
x++;
}
return 0;
}
I have a matrix written out here, and it appears on the screen fine but I want to put symbols into it. I can't seem to get the symbols (X) to appear in it. Sorry this is my first time working with arrays.
#include <iostream>
#define WIDTH 70
#define HEIGHT 20
using namespace std;
void main ()
{
char world[HEIGHT][WIDTH];
// draws matrix
for ( char i = 0; i < HEIGHT; i++ ) {
for ( char j = 0; j < WIDTH; j++ ) {
world[i][j] = '.';
cout << world[i][j];
}
cout << endl;
}
// 1st symbol at coordinates 1, 1
int x, y;
x = 1, y = 1;
world[x][y] = 'x';
cout << world[x][y];
// 2nd symbol at coordinates 2, 2
x = 2, y = 2;
world[x][y] = 'x';
cout << world[x][y];
//so on..
}
You have to first insert the xs, and then print the grid, not the other way around.
First of all, main should be declared as either int main() or int main(int argc, const char* argv[]); it certainly doesn't return void.
The standard output cout works sequentially. You just tell it what to print and it does it, one after the other. Consider the following:
int x = 1;
std::cout << x;
x = 2;
std::cout << x;
Because we changed the value of x and then passed it to cout again, it seems like you're expecting the output would be:
2
Actually the output is:
12
cout doesn't care that we passed it the same variable twice. It just prints each thing we give it one after the other.
You're going to have to set each element of the matrix you want to be an 'x', then loop through it again and print it all out.
The output is supposed to be in this fashion
I
IN
IND
INDI
INDIA
INDIA
INDI
IND
IN
I
I was wondering if there is anyway to make my program shorter.
Maybe use 2 while loops or 2 for loops instead of 4.
#include<iostream>
#include<conio.h>
using namespace std;
void main()
{
char a[]={'I','N','D','I','A'};
int r,c;
for(r=1;r<=5;r++)
{
for(c=0;c<r;c++)
cout<<a[c]<<" ";
cout<<endl;
}
for(r=5;r>=0;r--)
{
for(c=0;c<r;c++)
cout<<a[c]<<" ";
cout<<endl;
}
_getch();
}
This is nice and short, if you have C++11:
#include <stdio.h>
#include <initializer_list>
int main() {
for (int i : {1,3,5,7,9,9,7,5,3,1})
printf ("%.*s\n", i, "I N D I A") ;
}
I put it on ideone at this link.
The "cheating" ;-) but very short solution is
#include <cstdio>
int main() {
puts("I\nIN\nIND\nINDI\nINDIA\nINDIA\nINDI\nIND\nIN\nI\n");
}
Yet another solution based on TonyK's beautiful idea (works in C and C++):
#include <stdio.h>
int main() {
for (int i = 1; i < 11; ++i)
printf("%.*s\n", i < 6 ? i : 11 - i, "INDIA") ;
}
Update: (After TonyK's comment):
If you want spaces, then replace the printf above with this
printf("%.*s\n", 2*(i < 6 ? i : 11 - i), "I N D I A ");
Update: A solution with no loop:
#include <stdio.h>
int main(int i, char**) {
return printf("%.*s\n", i < 6 ? i : 11 - i, "INDIA") - 1 && main(i + 1, 0);
}
Detailed explanations for the benefit of beginners follow.
The most important point is that this is an "academic" exercise. The code is obscure and one should refrain from writing serious code like this.
The standard doesn't fix a particular signature for the function main. It only says that it must return an int and the compiler must accept two particular forms:
int main();
int main(int argc, char** argv);
I'm using the second one. The more popular form int main(int argc, char* argv[]) is equivalent to (2). Indeed, in general, an array of T used as a function argument is implicitly converted to pointer to T. In this example, char* argv[] is an array of pointer to char which becomes a pointer to pointer to char. The above program is not using the second parameter though (it's either ignored or given a NULL, i.e. 0 value).
Instead of doing a loop, the program calls main recursivelly. Actually, this is illegal but the major compilers turn a blind eye to it (at most GCC gives a warning if you ask it to be pedantic with option -Wpedantic). When the program is called from the command line with no arguments the operating system calls main passing 1 (which is 1 + the number of arguments) and a pointer to pointer to char which, as said, is ignored by this program. Hence, i = 1 at the first call.
Subsequently, main potentially calls itself incrementing i each time. In summary, main is sequentially called with i being 1, 2, 3, 4, ...
It's easy to see that for this sequence of values for i, the expression i < 6 ? i : 11 - i evaluates to 1, 2, 3, 4, 5, 5, 4, 3, 2, 1, 0, ... (This can also be expressed by 5.5 - abs(5.5 - i) as per Drew Dorman's solution but using the ternary operator avoids the need to #include <cmath> which saves a line ;-) These numbers are the quantity of characters of "INDIA" that should be displayed at each time.
The call printf("%.s\n", j, "INDIA"), where j = i < 6 ? i : 11 - i, displays the first j characters of "INDIA" followed by a new line and returns the number of characters effectivelly printed (including the new line), i.e. j + 1. Taking away 1 yields j.
We recall now an important point of the semantics of a && b, for integers a and b. If a == 0, then b is not evaluated. Otherwise, b is evaluated. For us, a = printf("%.*s\n", j, "INDIA") - 1 (i.e. a = j) and b = main(i + 1, 0).
Now, let's put these pieces together.
1. Initially, i = 1 and j = 1. The call to printf outputs "I\n" and returns 2. Taking away 1, gives a = 1 != 0. Therefore, b = main(2, 0) is evaluated (that is, main(2, 0) is called).
2. Well, for this second call to main, we have i = 2, j = 2, printf displays "IN\n", a = 2 != 0 and b = main(3, 0) is evaluated.
Repeating this argument, at each call to main we have:
3. i = 3, j = 3, "IND\n" is printed, a = 3 != 0 and main(4, 0) is called.
4. i = 4, j = 4, "INDI\n" is printed, a = 4 != 0 and main(5, 0) is called.
5. i = 5, j = 5, "INDIA\n" is printed, a = 5 != 0 and main(6, 0) is called.
6. i = 6, j = 5, "INDIA\n" is printed, a = 5 != 0 and main(7, 0) is called.
7. i = 7, j = 4, "INDI\n" is printed, a = 4 != 0 and main(8, 0) is called.
...
10. i = 10, j = 1, "I\n" is printed, a = 1 != 0 and main(11, 0) is called.
11. i = 11, j = 0, "\n" is printed, a = 0. Now main(12, 0) is not exectued and the main returns
Notice that main in step 1 calls main in step 2, which calls main in step 3, ... which calls main in step 11. Therefore, main in step 11 returns to main in step 10, which returns to main in step 9, ..., which returns to main in step 1. Finally, main in step 1 returns to the operating system.
If you are interested in even more obfuscated code, see present and past winners of the IOCCC. Enjoy!
Shorter.
#include <iostream>
#include <cmath>
int main()
{
char a[]={'I','N','D','I','A'};
for(int r=1;r<=10;r++)
{
int c = 5.5 - std::abs(5.5 - r);
std::cout << std::string(a, 0, c) << std::endl;
}
}
You could use substr.
Additionally, if you want to add spaces between the characters (as it's in your code right now) you could do that once for the whole word and then work on that modified string:
input = "INDIA";
modifiedInput = ...; // Figure out a way to get "I N D I A " here
for (int i = 1; ...)
cout << substr(modifiedInput, 0, 2 * i) << endl;
...
smth like this:
Anyway, 2 loops is hard thing)
int i=1;
bool ind=0;
for(r=0;r<10;r++)
{
if (ind == 0){
for(c=0;c<i;c++)
cout<<a[c]<<" ";
i++; }
if (ind == 1){
for(c=0;c<i;c++) cout<<a[c]<<" ";
i--;
}
if (i % 5 == 0) ind=1;
cout<<endl;
}
Here is an example using two loops:
#include<iostream>
using namespace std;
int main()
{
char a[]={'I','N','D','I','A'};
int arr_length=5;
bool two_times=false;//if you want to output INDIA twice
for(int i=2*arr_length-1; i>=1; i--)
{
for(int j=0; j<arr_length-abs(arr_length-i); j++)
{
cout << a[j];
}
if(i==arr_length && !two_times){ two_times=true; i++;}//if you want to output INDIA twice
cout << endl;
}
return 0;
}
Output:
I
IN
IND
INDI
INDIA
INDIA
INDI
IND
IN
I
You can do like this:
void main()
{
char a[]={'I','N','D','I','A'};
int r,c;
int x = 0; // initially x is 0
int l = 5;
for(r=1;r<= 11 ;r++) //loop runs for 5*2 + 1 times
{
if(r>=6){
x = l - r % 6; //set x to remaining length
l--;
}
for(c=0;c< (r % 6) + x; c++) // add x
cout<<a[c]<<" ";
cout<<endl;
}
}
and you can find it working here.
Another solution based on TonyK's answer. You can enter your desired string.
#include <stdio.h>
#include <string.h>
int main() {
char s[100];
gets(s);
for (int i = 0; i < strlen(s) * 2; i++)
printf ("%.*s\n", i < strlen(s) ? i + 1: 2 * strlen(s) - i, s);
}
http://ideone.com/Zsg5Lu