When the calculator shows \(X\), multiplying by \(A_i\) makes it show:

• \(1\) if \((X\times A_i)>10^K-1\), and
• \((X\times A_i)\) otherwise.

It is sufficient to perform this for \(i=1,2,\ldots,K\), and find the integer finally shown.

However, \((X\times A_i)\) can be up to \((10^{18}-1)^2\), causing an overflow error if we compute using \(64\)-bit integer type, so we need a trick. For example, one can utilize the following property: for \(Y=10^K-1\), \((X\times A_i)>Y\) if and only if \(X> \left\lfloor \frac{Y}{A_i}\right\rfloor\), where \(\left\lfloor \frac{Y}{A_i}\right\rfloor\) denotes the maximum integer not exceeding \( \frac{Y}{A_i}\).

In this case, \(X\) and \(\left\lfloor \frac{Y}{A_i}\right\rfloor\) are both \(10^{18}\), and the latter can be found as an integer division, so it can be computed using \(64\)-bit integer type without an overflow. Alternatively, one can use multi-precision integers (bigints).

Sample code in C++:

#include <bits/stdc++.h>
using namespace std;

int main(void){
	int n,k;
	long long a,y=1,x=1;

	cin>>n>>k;

	for(int i=0;i<k;i++)y*=10;
	y--;

	for(int i=0;i<n;i++){
		cin>>a;
		if(x>(y/a))x=1;
		else x*=a;
	}

	cout<<x<<endl;
	return 0;
}

Sample code in Python:

n,k=map(int, input().split())
a=list(map(int, input().split()))

x,y=1,10**k-1

for i in range(n):
   x*=a[i]
   if x>y:
       x=1

print(x)