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# Queries on sum of odd number digit sums of all the factors of a number

Given Q queries. Each query contain a positive integer n. The task is to output the sum of sum of odd number digit contained in all the divisors of n.
Examples :

Input : Q = 2, n1 = 10, n2 = 36
Output : 7 18
For Query1,
Divisors of 10 are 1, 2, 5, 10.
Sum of odd digits in 1 is 1, in 2 is 0, in 5 is 5, in 10 is 1.
So, sum became 7.
For Query 2,
Divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.
Sum of odd digits in 1 is 1, in 2 is 0, in 3 is 3, in 4 is 0,
in 6 is 0, in 9 is 9, in 12 is 1, in 18 is 1, in 36 is 3.
So, sum became 18.

Approach:
The idea is to precompute the sum of odd number digit of all the numbers. Also, we can you use the sum of odd number digit of the previous number to compute the sum of odd number digit of the current number.
For example, to compute the sum of odd number digit of “123”, we can use the sum of odd number digit of “12” and “3”. Therefore, the sum of odd digit of “123” = sum of odd digit of “12” + add the last digit if it is odd (i.e 3).
Now, to find the sum of the sum of odd number digit of the factors, we can you use the jump phenomenon of Sieve of Eratosthenes. So, for all possible factors, add their contribution to its multiples.
For example, for 1 as the factor, add 1 (because 1 have only 1 odd digit) to all of its multiple.
for 2 as the factor, add 0 to all the multiples of 2 i.e 2, 4, 8, …
for 3 as the factor, add 1 to all the multiples of 3 i.e 3, 6, 9, …..

Below is the implementation of this approach:

## C++

 `// CPP Program to answer queries on sum``// of sum of odd number digits of all``// the factors of a number``#include ``using` `namespace` `std;``#define N 1000005` `// finding sum of odd digit number in each integer.``void` `sumOddDigit(``int` `digitSum[])``{``    ``// for each number``    ``for` `(``int` `i = 1; i < N; i++) {` `        ``// using previous number sum, finding``        ``// the current number num of odd digit``        ``// also, adding last digit if it is odd.``        ``digitSum[i] = digitSum[i / 10] + (i & 1) * (i % 10);``    ``}``}` `// finding sum of sum of odd digit of all``// the factors of a number.``void` `sumFactor(``int` `digitSum[], ``int` `factorDigitSum[])``{``    ``// for each possible factor``    ``for` `(``int` `i = 1; i < N; i++) {``        ``for` `(``int` `j = i; j < N; j += i) {` `            ``// adding the contribution.``            ``factorDigitSum[j] += digitSum[i];``        ``}``    ``}``}` `// Wrapper function``void` `wrapper(``int` `q, ``int` `n[])``{``    ``int` `digitSum[N];``    ``int` `factorDigitSum[N];` `    ``sumOddDigit(digitSum);``    ``sumFactor(digitSum, factorDigitSum);` `    ``for` `(``int` `i = 0; i < q; i++)``        ``cout << factorDigitSum[n[i]] << ``" "``;``}` `// Driver Program``int` `main()``{``    ``int` `q = 2;``    ``int` `n[] = { 10, 36 };` `    ``wrapper(q, n);``    ``return` `0;``}`

## Java

 `// Java Program to answer queries``// on sum of sum of odd number ``// digits of all the factors of``// a number``class` `GFG``{``    ``static` `int` `N = ``1000005``;``    ` `    ``// finding sum of odd digit``    ``// number in each integer.``    ``static` `void` `sumOddDigit(``int` `digitSum[])``    ``{``        ` `        ``// for each number``        ``for` `(``int` `i = ``1``; i < N; i++)``        ``{``    ` `            ``// using previous number sum,``            ``// finding the current number``            ``// num of odd digit also,``            ``// adding last digit if it``            ``// is odd.``            ``digitSum[i] = digitSum[i / ``10``] +``                         ``(i & ``1``) * (i % ``10``);``        ``}``    ``}``    ` `    ``// finding sum of sum of odd digit``    ``// of all the factors of a number.``    ``static` `void` `sumFactor(``int` `digitSum[],``                    ``int` `factorDigitSum[])``    ``{``        ` `        ``// for each possible factor``        ``for` `(``int` `i = ``1``; i < N; i++)``        ``{``            ``for` `(``int` `j = i; j < N; j += i)``            ``{``                ``// adding the contribution.``                ``factorDigitSum[j] += digitSum[i];``            ``}``        ``}``    ``}``    ` `    ``// Wrapper function``    ``static` `void` `wrapper(``int` `q, ``int` `n[])``    ``{``        ``int` `digitSum[] = ``new` `int``[N];``        ``int` `factorDigitSum[] = ``new` `int``[N];``    ` `        ``sumOddDigit(digitSum);``        ``sumFactor(digitSum, factorDigitSum);``    ` `        ``for` `(``int` `i = ``0``; i < q; i++)``            ``System.out.print(factorDigitSum[n[i]]``                                          ``+ ``" "``);``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `q = ``2``;``        ``int` `n[] = ``new` `int``[]{``10``, ``36``};``    ` `        ``wrapper(q, n);``        ` `    ``}``}` `// This code is contributed by Sam007`

## Python3

 `# Python Program to answer queries``# on sum of sum of odd number``# digits of all the factors``# of a number``N ``=` `100``digitSum ``=` `[``0``] ``*` `N``factorDigitSum ``=` `[``0``] ``*` `N` `# finding sum of odd digit``# number in each integer.``def` `sumOddDigit() :``    ``global` `N,digitSum,factorDigitSum``    ` `    ``# for each number``    ``for` `i ``in` `range``(``1``, N) :``        ` `        ``# using previous number``        ``# sum, finding the current``        ``# number num of odd digit``        ``# also, adding last digit``        ``# if it is odd.``        ``digitSum[i] ``=` `(digitSum[``int``(i ``/` `10``)]``                    ``+` `int``(i & ``1``) ``*` `(i ``%` `10``))` `# finding sum of sum of``# odd digit of all the``# factors of a number.``def` `sumFactor() :``    ``global` `N,digitSum,factorDigitSum``    ``j ``=` `0``    ` `    ``# for each possible factor``    ``for` `i ``in` `range``(``1``, N) :``        ``j ``=` `i``        ``while` `(j < N) :``            ` `            ``# adding the contribution.``            ``factorDigitSum[j] ``=` `(factorDigitSum[j]``                                   ``+` `digitSum[i])``            ``j ``=` `j ``+` `i` `# Wrapper def``def` `wrapper(q, n) :` `    ``global` `N,digitSum,factorDigitSum``    ` `    ``for` `i ``in` `range``(``0``, N) :    ``        ``digitSum[i] ``=` `0``        ``factorDigitSum[i] ``=` `0``    ` `    ``sumOddDigit()``    ``sumFactor()` `    ``for` `i ``in` `range``(``0``, q) :``        ``print` `(``"{} "``.``        ``format``(factorDigitSum[n[i]]), end ``=` `"")` `# Driver Code``q ``=` `2``n ``=` `[ ``10``, ``36` `]``wrapper(q, n)` `# This code is contributed by``# Manish Shaw(manishshaw1)`

## C#

 `// C# Program to answer queries on sum``// of sum of odd number digits of all``// the factors of a number``using` `System;` `class` `GFG {``    ` `    ``static` `int` `N = 1000005;``    ` `    ``// finding sum of odd digit number in``    ``// each integer.``    ``static` `void` `sumOddDigit(``int` `[]digitSum)``    ``{``        ` `        ``// for each number``        ``for` `(``int` `i = 1; i < N; i++) {``    ` `            ``// using previous number sum,``            ``// finding the current number``            ``// num of odd digit also,``            ``// adding last digit if it``            ``// is odd.``            ``digitSum[i] = digitSum[i / 10]``                     ``+ (i & 1) * (i % 10);``        ``}``    ``}``    ` `    ``// finding sum of sum of odd digit``    ``// of all the factors of a number.``    ``static` `void` `sumFactor(``int` `[]digitSum,``                     ``int` `[]factorDigitSum)``    ``{``        ` `        ``// for each possible factor``        ``for` `(``int` `i = 1; i < N; i++) {``            ``for` `(``int` `j = i; j < N; j += i)``            ``{``                ``// adding the contribution.``                ``factorDigitSum[j] += digitSum[i];``            ``}``        ``}``    ``}``    ` `    ``// Wrapper function``    ``static` `void` `wrapper(``int` `q, ``int` `[]n)``    ``{``        ``int` `[]digitSum = ``new` `int``[N];``        ``int` `[]factorDigitSum = ``new` `int``[N];``    ` `        ``sumOddDigit(digitSum);``        ``sumFactor(digitSum, factorDigitSum);``    ` `        ``for` `(``int` `i = 0; i < q; i++)``            ``Console.Write(factorDigitSum[n[i]]``                                       ``+ ``" "``);``    ``}``        ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `q = 2;``        ``int` `[]n = ``new` `int``[]{ 10, 36 };``    ` `        ``wrapper(q, n);``    ``}``}` `// This code is contributed by Sam007.`

## PHP

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## Javascript

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Output :

`7 18`

Time complexity: O(q+N2)
Auxiliary space: O(N)

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