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Product of factors of number in C++
Given a number n we have to find its all factors and find the product of those factors and return the result, i.e, the product of factors of a number. Factors of a number are those numbers which can divide the number completely including 1. Like factors of 6 are − 1, 2, 3, 6.
Now according to the task we have to find the product all the factors of the number.
Input − n = 18
Output − 5832
Explanation − 1 * 2 * 3 * 6 * 9 * 18 = 5832
Input − n = 9
Output − 27
Explanation − 1 * 3 * 9 = 27
Approach used below is as follows to solve the problem −
Take the input num .
Loop from i = 1 till i*i<=num
Check if the num%i==0 then, check
If num%i == i then set the value of product = (product*i)%1e7
Else Set product as (product * i) % MAX and Set product as (product * num / i) % MAX.
Return the product.
Algorithm
Start In Function long long productfactor(int num) Step 1→ Declare and Initialize product as 1 Step 2→ For i = 1 and i * i <= num and i++ If num % i == 0 then, If num / i == i then, Set product as (product * i) % MAX Else Set product as (product * i) % MAX Set product as (product * num / i) % MAX Step 3→ Return product In Function int main() Step 1→ Declare and initialize n as 9 Step 2→ Print the result productfactor(n) Stop
Example
#include <stdio.h>
#define MAX 1000000000
// find the product of the factors
long long productfactor(int num){
long long product = 1;
for (int i = 1; i * i <= num; i++){
if (num % i == 0){
//equal factors should be multiplied only once
if (num / i == i)
product = (product * i) % MAX;
// Else multiply both
else {
product = (product * i) % MAX;
product = (product * num / i) % MAX;
}
}
}
return product;
}
int main(){
int n = 9;
printf("%lld\n", productfactor(n));
return 0;
}Output
If run the above code it will generate the following output −
27
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