Monday 7 March 2016

Solution of URI 1075 :: Remaining 2


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 N,a;
    scanf("%d",&N);
    for(a=2;a<=10000;a=a+N)
        printf ("%d\n",a);
    return 0;
}

3 comments:

  1. #include
    #include
    #include

    int main()
    {
    int i,N;
    scanf("%d",&N);
    for(i=1;i<=1000;i++)
    {
    if(i%N==2) printf("%d\n", i);
    }


    return 0;
    }

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