In this HackerEarth Sequences problem solution you are given two integers X and Y. your task is to determine the sequence of real numbers with the minimum size such that the sum and product of the numbers in the sequence are X and Y respectively. Also, print the size of the sequence only. if no such sequences exist, then print -1.

## HackerEarth Sequences problem solution.

```#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back

bool check(int n,int y,int x){
double v = temp(n,x);
return (v >= y );
}

double temp(int n,int x){
return pow(x*1.0/n, n);
}

int main() {
int t;
cin>>t;
while(t--) {
int x,y;
cin>>x>>y;
int high = (int)floor(x/exp(1.0));
int low = 1;
if (x == y) cout<<1<<endl;
else if (!check(high,y,x)) cout<<-1<<endl;
else {
while(high - low > 1){
int mid = (high+low)/2;
if (check(mid,y,x)) high = mid;
else low = mid;
}
cout<<high<<endl;
}
}
return 0;
}```

### Second solution

```#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 1e6 + 17;
int main(){
ios::sync_with_stdio(0), cin.tie(0);
int t;
cin >> t;
while(t--){
int x, y;
cin >> x >> y;
bool done = 0;
if(x == y)
cout << "1\n";
else{
for(int i = 2; !done && i < 100; i++)
if(pow(x / double(i), i) >= y){
cout << i << '\n';
done = 1;
}
if(!done)
cout << "-1\n";
}
}
}```