In this HackerEarth Perfect Subarray problem solution, You are given an array A of N integers. You need to count the subarrays which have the sum of all elements in it a perfect square.

## HackerEarth Perfect Subarray problem solution.

```#include<bits/stdc++.h>
#define LL long long int
#define M 1000000007
#define endl "\n"
#define eps 0.00000001
LL pow(LL a,LL b,LL m){LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}
LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}
LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}
using namespace std;
LL sum[100001];
int a[100001];
bool f(LL n){
LL temp = sqrt(n);
if(temp * temp == n){
return 1;
}
++temp;
if(temp * temp == n){
return 1;
}
return 0;
}
int main()
{
ios_base::sync_with_stdio(0);
int n;
cin >> n;
assert(n >= 1 && n <= 3000);
for(int i = 1; i <= n; i++){
cin >> a[i];
assert(a[i] >= 1 && a[i] <= 1000000);
sum[i] = sum[i - 1] + a[i];
}
int ans = 0;
for(int i = 1; i <= n; i++){
for(int j = i; j <= n; j++){
if(f(sum[j] - sum[i - 1]))
++ans;
}
}
cout << ans;
}```

### Second solution

```#include <bits/stdc++.h>

using namespace std;

const int N = 5E3 + 5;
int a[N];

int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
cin >> n;
for(int i = 0; i < n; i ++)
cin >> a[i];
int ans = 0;
for(int i = 0; i < n; i ++) {
double total = 0;
for(int j = i; j < n; j ++) {
total += a[j];
double chkr = total;
chkr = sqrt(chkr);
if(chkr == floor(chkr))
ans ++;
}
}
cout << ans;
return 0;
}```