Suppose we have an array of n elements called A. We have another number k. Our task is to find last k digits of the product of elements in the array A. Suppose the A = [15, 22, 13, 19, 17], then the product is 1385670, the last k = 3 digits are 670.
To solve this problem, we will multiply the numbers under modulo 10k.
Example
#include<iostream>
#include<cmath>
using namespace std;
int displayLastKNumbers(int array[], int n, int k) {
int mod = (int)pow(10, k);
int mul = array[0] % mod;
for (int i = 1; i < n; i++) {
array[i] = array[i] % mod;
mul = (array[i] * mul) % mod;
}
return mul;
}
int main() {
int a[] = {15, 22, 13, 19, 17};
int k = 3;
int n = sizeof(a) / sizeof(a[0]);
cout <<"Last K digits are: " << displayLastKNumbers(a, n, k);
}Output
Last K digits are: 670