# URI / BEE CROWD 1095 - Sequence IJ 1 Solution in C,C++,Python | URI - BEECROWD - BEE 1095 Solution in C,C++,Python

beecrowd | 1095

# Adapted by Neilor Tonin, URI  Brazil

Timelimit: 1

URI / BEE CROWD 1095 - Sequence IJ 1 Solution in C,C++,Python | URI - BEECROWD - BEE 1095 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=1 J=60I=4 J=55I=7 J=50...I=? J=0

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.

Start with i =1 and j =60 . For every iteration, increment i by 3 and decrement by 4.
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 1095 Solve in C :

```#include<stdio.h>

int main()
{
int i,j;
for(i=1,j=60;i<15,j>=0;i+=3,j-=5){

printf("I=%d J=%d\n",i,j);
}

}```

### 1 Response to URI / BEE CROWD 1095 - Sequence IJ 1 Solution in C,C++,Python | URI - BEECROWD - BEE 1095 Solution in C,C++,Python

1. for j in range(60,-1,-5):
if j==60: i =1
print(f'I={i} J={j}')
i+=3
#This is my Logic