我不知道如何修复我的代码，它不能正常工作

Question

I have a question about c++ and a problem which I have to solve. Actually, I wrote a code but the online judge gives me 70 from 100 and says that my code is not working correctly for 3 test cases. I can't find the bug to correct it, I would appreciate if anyone can help me. Thanks

I want to print the output for every 2 natural numbers a, b which make a/b fraction. for example for the number (2) it should print 1 1 in the output because we don't consider (0,0) and the second point which we reach in the coordinate on the spiral path is (1,1). the input should be a natural number and the output should be vertical and horizontal coordinate.

for example, the input is 12 and its output is 2 2. more information is here in these photos:

#include<iostream>

using namespace std;

int main()
{
    int i,j,x,y,count;
    unsigned int n;
    x=0;
    y=0;
    count=0;
    cin>>n;
    for(i=1;i<=30000;i++)
    {
        if(i%2!=0)
        {
            for(j=0;j<i;j++)
            {
                x++;
                count++;
                if(count==n)
                {
                    cout<<x<<" "<<y;
                }
            }
            for(j=0;j<i;j++)
            {
                y++;
                count++;
                if(count==n)
                {
                    cout<<x<<" "<<y;
                }
            }
        }
        else
        if(i%2==0)
        {
            for(j=0;j<i;j++)
            {
                x--;
                count++;
                if(count==n)
                {
                    cout<<x<<" "<<y;
                }
            }
            for(j=0;j<i;j++)
            {
                y--;
                count++;
                if(count==n)
                {
                    cout<<x<<" "<<y;
                }
            }
        }
    }
}

Answer 1

Your algorithm maybe a little bit too complicated. Especially running loops always 30000 times , even while not necessary, is wasting time somehow.

And, if you want to request numbers for input > 30000, it will also not work.

The loop should be rewritten as: for (i = 1; count <= n; i++) . Then is should be better.

---

If you just want to have xycoordinates for a spiral pattern, you may do the following design and then implement it. Looking at the picture, we see the follwoing:

• We will have 2 times (for 2 directions) the same distance or length for a line
• Then, if we work on one line with a given distance or length, we always add the same delta value of wither 1 or -1 to a coordinate.
• After we have worked one 1 line, we will add delta values to teh other coordinate. If it was x before than it will be y now and vice versa
• To do that, we can store the coordinates in an array with length 2. Index 0 is for x and index 1 is for y.
• So, if we want to address either x or y, we can use a selector, a index, that will be either 0 or 1.
• We can toggle this selector after one distance or one line has been done
• After we have worked on 2 distances or 2 lines,we needto change the delta values, becuase from then on coordinated need to be decremented now. The same for vice versa.
• Additionally, after 2 distances or lines have been done, we will increase the distance or length to be used next time by 1.

So, look at the pricture and implement accordingly.

Please see one of many solution proposals:

#include <iostream>

int main() {

    // Get user input. How many steps shall we go in our spiral
    unsigned int numberOfSteps{};
    std::cin >> numberOfSteps;

    int coordinate[2]{};            // Here we will store the xy coordinate. Index 0 = x, index 1 = y
    int xySelector{};               // This will select either index 0 or 1 , so, either x or y

    int delta{ 1 };                 // This will be added to a coordinate part, to get the next point of the line

    int distance{ 1 };              // Length of line. Will be used 2 times, then incremented
    int distanceCounter{ 1 };       // Down counter for distance
    int doSameDistanceCounter{ 2 }; // We will do always 2 lines with the same distance/length

    // Do the game for the requested number of steps
    for (int i{}; i < numberOfSteps; ++i) {

        coordinate[xySelector] += delta;    // Calculate next point of line

        --distanceCounter;                  // Keep track of length of line
        if (distanceCounter == 0) {         // Did we calculate one complete line?
            xySelector = 1-xySelector;      // Yes, then now use the other coordinate 
            --doSameDistanceCounter;        // Keep tract of number of same line length to use
            if (doSameDistanceCounter == 0) {   // If we did 2 lines with the same length
                doSameDistanceCounter = 2;      // Reset this counter
                ++distance;                     // Frome now one the length of line will be 1 more
                delta = -delta;                 // And the delta changes, fo increment or decrement operation changes
            }
            distanceCounter = distance;     // Reset distance/length counter to current needed distance
        }
    } // Show result
    std::cout << coordinate[0] << ' ' << coordinate[1] << '\n';
}

I tested it in release mode with 10M and the result was ultra fast.