QUESTION

Fredo is assigned a task today. He is given an array A containing N integers. His task is to update all elements of array to some minimum value x that is, A[i]=x; i<=1<=N such that product of all elements of this new array is strictly greater than the product of all elements of the initial array. \n\nNote that x should be as minimum as possible such that it meets the given condition. Help him find the value of x.\n\nInput Format:\nThe first line consists of an integer N denoting the number of elements in the array.\nThe next line consists of N space separated integers, denoting the array elements.\n\nOutput Format:\nThe only line of output consists of value of x\n\nInput Constraints:\n \n1<=N<=10 to the power 5\n1<=A[i]<=10 to the power 10\nA[i] denoting an array element.\n\nInitial array product = 4*2*1*10*6=480.\nIf all elements becomes 4 then product=4*4*4*4*4=1024.Had all elements become 3\n3, product would be =243 which is less than 480\n\nHence, value of \nx is 4.

“TESTCASE_1”: “5\n4 2 1 10 6\n###—###SEPERATOR—###—\n4”, “TESTCASE_2”: “5\n27 253 316 3830 2384\n###—###SEPERATOR—###—\n456”, “TESTCASE_3”: “0\n###—###SEPERATOR—###—\n0”, “TESTCASE_4”: “0\n###—###SEPERATOR—###—\n0”, “TESTCASE_5”: “0\n###—###SEPERATOR—###—\n0

ANSWER

#include <stdio.h>
#include <math.h>
int main()
{
    int n;
    scanf("%d",&n);
    unsigned long long int a,q=1;
    long double p=1.0;
  int i;
    for(i=1;i<=n;i++){
        scanf("%lld",&a);
        p*=pow(a,1.0/n);
    }
     int r=floor(p);
    r++;
    printf("%d\n",r);
    return 0;
}