Monday, 7 March 2016

Solution of URI 1097 :: Sequence IJ 3

   , ,  2 comments

Before seeing the solution make sure that you tried enough. Don’t paste the whole code, just find out the logic. If you stuck in trouble, just inform me on comment.

/**Bismillahir Rahmanir Rahim.**/

#include <stdio.h>
int main()
{
    int I,J,a=7,b;
    for(I=1;I<=9;I=I+2)
    {
        for(b=1,J=a;b<=3;J--,b++)
            printf("I=%d J=%d\n", I, J);
        a=a+2;
    }
    return 0;
}

2 comments:

  1. Unknown24 August 2019 at 21:59

    #include

    int main(int argc, char const *argv[])
    {
    int i, j, k = 0;

    for (i = 1; i <= 9; i += 2, k += 2)
    {
    for (j = 7 + k; j >= 5 + k; j--)
    {
    printf("I=%d J=%d\n", i, j);
    }

    }

    return 0;
    }

    I think this is easier to understand

    ReplyDelete
  2. Unknown4 April 2020 at 14:16

    #include
    int main()
    {
    int I=1,J=7;

    while(I!=11)
    {

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

    J=J-1;
    printf("I=%d J=%d\n",I,J);
    J=J-1;
    printf("I=%d J=%d\n",I,J);
    I=I+2;
    J=J+4;




    }
    }

    ReplyDelete

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