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Product of Digits

TIME LIMIT = 1 SEC.

The natural numbers begin from 1 and go on up to ∞.
Given N indices, find the product of the digits that lie on those indices.
NOTE:
- Consider that the natural numbers are zero-indexed i.e. the index of the first number in the set of natural numbers is 0, the index of the second number is 1 and so on.
- Consider that all the natural numbers are written like a string '1234567891011121314….'
Now, let D be a function that takes an index as it's input and returns the digit that lies on that index.
D(0)=1;D(1)=2;D(2)=3 and so on…

Input	Output
First line will contain N, number of indices. Second line will contain the Ni indices, each separated by a space.	Print the product of the digits that lie on those indices.

Constraints

1 ≤ N ≤ 10
1 ≤ Ni ≤ 1000

Example Test Case

Input	Output
5 1 2 3 4 5	720

Solution

#include <iostream> #include<set> #include<string> using namespace std; int main() { int n,i,temp; cin>>n; long product=1; multiset<int> indices; //stores the indices multiset<int>::reverse_iterator it; //to interate in reverse order string digits; //to store number as string //take indices for(i=0;i<n;i++) { cin>>temp; indices.insert(temp); } //start from last it = indices.rbegin(); //append each number to string. for(i=1;i<=*it+1;i++) digits += to_string(i); //get the digit at index and multiply with product. for(it=indices.rbegin();it!=indices.rend();++it) product *= (digits[*it]-48); //convert to int and multiply cout<<product; //display product. return 0; }