Can anyone help me debugging a simple program?

Question

1.00/5 (2 votes)

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This is the problem:
1098 - Sequence IJ 4 - beecrowd[^]

What I have tried:

#include<stdio.h>
int main(){
double i,j;


for(i=0; i<=2; i=i+0.2){
    for(j=1; j<=3; j++){

        if (i==0){
            printf("%.0lf %.0lf\n", i, j+i);
        }else if(i==1){
        printf("%.0lf %.0lf\n", i, j+i);
        }else if(i==2){
        printf("%.0lf %.0lf\n", i, j+i);
        }else{
        printf("%.1lf %.1lf\n", i, j+i);
        }
    }
}
return 0;
}

Posted 26-Oct-21 15:30pm

Jahirul Sarker

Updated 26-Oct-21 19:14pm

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3 solutions

Solution 4

Here's the solution.

#include<stdio.h>

int main(){
double i,j;


for(i=0; i<=2; i=i+0.2){
    for(j=1; j<=3; j++){

        if (i<0.1){
            printf("I=%.0lf J=%.0lf\n", i, j+i);
        }else if(i>0.9&&i<1.1){
        printf("I=%.0lf J=%.0lf\n", i, j+i);
        }else if(i>1.8&&i<2.1){
        printf("I=%.0lf J=%.0lf\n", i, j+i);
        }else{
        printf("I=%.1lf J=%.1lf\n", i, j+i);
        }
    }
}
return 0;
}

Posted 26-Oct-21 19:14pm

Jahirul Sarker

Comments

Rick York 27-Oct-21 17:27pm

That might work but it has serious limitations. I have revised my solution to show you what it should be more like.

Solution 1

by adding .2 to i you are never going to match your if statements, it will always default to the last else statement.

Posted 26-Oct-21 15:55pm

The Other John Ingram

Comments

Jahirul Sarker 26-Oct-21 23:09pm

What should I do?

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Rick York · Accepted Answer · 2021-10-26T15:59:00

The answer is the same as the last time you asked the question : 'Else if' is not working[^]

Using the CloseEnough function from before, here is how this could look :

C++

int main()
{
    double i,j;

    for( i=0; i<=2; i += 0.2 )
    {
        for( j=1; j <= 3; ++j )
        {
            if( CloseEnough( i, 0 ) )
            {
                printf("%3.1f %3.1f\n", i, j+i);
            }
            else if( CloseEnough( i, 1 ) )
            {
                printf("%3.1f %3.1f\n", i, j+i);
            }
            else
            {
                printf("%3.1f %3.1f\n", i, j+i);
            }
        }
    }
    return 0;
}