Algorithm of 12043 - Dvisor

Problem Description Link

Algorithm:

In this problem if you use more loop in main program to determine the number of divisor and sum of divisor may be you get time limit exit .
so before test case you must generate two array for divisor and sum_of_divisor and and finally use a loop l
i=a to b and check that if any value i mod k is zero then add div+=divisor[i] and sum+=sum_of_divisor[i];
To generate divisor[] and sum_of_divisor[] you can use siv method
that method is bellow :
at first you must set each element 1 in divisor[] and sum_of_divisor[] after then
    for(i=2;i<=max;i++)// max is array size
    {
        dividor[i]+=1;
        sumdividor[i]+=i;
        for(j=i+i;j<=max;j=j+i)
        {
           dividor[j]+=1;
           sumdividor[j]+=i;
        }
    }


in min program you need only check

    if(j%k==0)//where j=a to b
                {
                    div+=dividor[j];
                    sum+=sumdividor[j];
                }
div and sum are result.