UVA 11879 - Multiple of 17

Mon, 29 Jan 2018 13:48:00 +0000

11879 - Multiple of 17

Problem Link: https://uva.onlinejudge.org/external/118/11879.pdf

Algorithm: This problem is quite easy. In this, there will be a number which is greater at most 100 digits. So, Its not possible to store it in “long long int” in c++.

We will write a program which will take input a string. If the string contains only 0 then the program will be break, otherwise it will enter in another loop. Before entering I will declare two integer variables ‘reminder’ and ‘i’. I will initialize the reminder variable to 0;

UVA 11876 - N + NOD (N)

Mon, 29 Jan 2018 13:23:00 +0000

Description:

For a number N, first have to calculate N-1+divisors(N-1) & store it in array (e.g. arr[i]=arr[i-1]+divisors(arr[i-1])). Make a sequence from 1 to until arr[i] becomes greater than 1000000. Then you will be given two numbers A & B, and you will have to find the number of integers from the above sequence which lies within the range [A,B].

Hints = No critical case. Easy problem.

Algorithm:

Step 1: First find the prime numbers from 1 to 1000000 and store them in an array (was not sure about the range, thats why I calculated till 1000000) . Step 2: Calculate number of divisors for each numbers from 1 to 1000000 and store divisors in each array. (e.g. divisors of 6 are 1,2,3,6. Total 4 divisors. So divisor[6]=4. ) Step 3: To find the divisors of each number, I took help of prime numbers. There is a good tutorial in lightoj.com to find divisors using primes. Link is give below: http://lightoj.com/article_show.php?article=1003

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UVA 11879 - Multiple of 17

11879 - Multiple of 17

UVA 11876 - N + NOD (N)

Description:

Algorithm: