# URI / BEE CROWD 1098 - Sequence IJ 4 Solution in C,C++,Python | URI - BEECROWD - BEE 1098 Solution in C,C++,Python

beecrowd | 1098

# Adapted by Neilor Tonin, URI  Brazil

Timelimit: 1

URI / BEE CROWD 1098 - Sequence IJ 4 Solution in C,C++,Python | URI - BEECROWD - BEE 1098 Solution in C,C++,Python:

Make a program that prints the sequence like the following example.

## Input

This problem doesn't have input.

## Output

Print the sequence like the example below.

 Input Sample Output Sample I=0 J=1I=0 J=2I=0 J=3I=0.2 J=1.2I=0.2 J=2.2I=0.2 J=3.2.....I=2 J=?I=2 J=?I=2 J=?

Demonstration:

This program does not take user inputs . Then check the conditions for each test case contains two integer number i and j .Then calculating the sequence by using conditional loop.

Finally print the result.

N.B: Don't copy paste the code as same. Just try to understand it and try yourself.

## URI / BEECROWD Online Judge 1098 Solve in C :

```#include <stdio.h>

int main(){
float i = 0;
float j = 1;

while(i <= 2.2){
if((i>0 && i<1)|| (i>1 && i<2) || (i>2.1 && i<=2.2) )
{
printf("I=%0.1f J=%0.1f\n",i,j+i);
printf("I=%0.1f J=%0.1f\n",i,j+1+i);
printf("I=%0.1f J=%0.1f\n",i,j+2+i);
}
else
{
printf("I=%d J=%d\n",(int)i,(int)j+(int)i);
printf("I=%d J=%d\n",(int)i,(int)j+1+(int)i);
printf("I=%d J=%d\n",(int)i,(int)j+2+(int)i);
}

i+= 0.2;
}
return 0;
}```