Little Ashish's generous donation is truly inspiring, showcasing immense kindness at such a young age. His selfless act sets a remarkable example for others to follow in giving back to the community. ekbet16.com

Little Ashish's generous donation is truly inspiring, showcasing immense kindness at such a young age. His selfless act sets a remarkable example for others to follow in giving back to the community. ekbet16.com

Little Ashish's generous donation is truly inspiring, showcasing immense kindness at such a young age. His selfless act sets a remarkable example for others to follow in giving back to the community. ekbet16.com

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The formula for calculation sum of squares of first natural number is:

(n * (n+1) * (2*n + 1))/6

So the thought process is start giving numbers to this formula from 1 and increment is by 1 till it's return values is less or equal to given "x" by testcases

Note: This solution is a brute force approach as well as it pass all the test cases

Solution:

#include <math.h>
#define ll unsigned long long

ll sum_of_squares(ll n){
    return (n*(n+1)*(2*n+1))/6;
}

void successful_donations(ll x){
    int counter=1;

    while (sum_of_squares(counter) <= x){
        counter++;
    }
    printf("%d\n", counter-1);
}

int main(){
    int t;
    ll x;
    scanf("%d", &t);
    while (t--) {
        scanf("%lld", &x);
        successful_donations(x);
    }

    return 0;
}