Here is the question:
Comparing two numbers written in index form like 2^11 and 3^7 is not difficult, as any calculator would confirm that 2^11=2048<3^7=2187.
However, confirming that 632382^518061>519432^525806 would be much more difficult, as both numbers contain over three million digits.
You are given N base exponent pairs, each forming a large number you have to find the Kth smallest number of them. K is 1−indexed.
Input Format First line containts an integer N, number of base exponent pairs. Followed by N lines each have two space separated integers B and E, representing base and exponent. Last line contains an integer K, where K<=N Constraints 1≤N≤105 1≤K≤N 1≤B≤109 1≤E≤109 No two numbers are equal.
Here is my code:
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int N,i = 0,k,j=0,x,m;
long long int *arr,*arr2,*arr3;
cin >> N;
arr = (long long int *)malloc(sizeof(long long int)*2*N);
arr2 = (long long int *)calloc(N,sizeof(long long int));
arr3 = (long long int *)calloc(N,sizeof(long long int));
x = 2*N;
while(x>0)
{
cin >> arr[i];
i++;
x--;
}
cin >> k;
for(i=0;i<2*N;i+=2)
{
arr2[j] = pow(arr[i],arr[i+1]);
j++;
}
arr3 = arr2;
sort(arr2,arr2+N);
for(i=0;i<N;i++)
{
if(arr3[i] == arr2[k-1])
{
m = i;
break;
}
}
cout << arr[2*m] << " " << arr[2*m + 1];
return 0;
}
The program works for small numbers only, can't make it work for large numbers. what to do?