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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;
}

Given the sum of squares formula, the number of days to spend x candies is roughly the cube root of 3x. So I back off a few days from that estimate to be conservative and then iterate through until I find the right value.

def solve(candies):
    day = math.floor(pow(3*candies,1/3))-3
    while day*(day+1)*(2*day+1) <= 6*candies:
        day += 1
    return day-1

c++ almost complete all test cases

int solve(long x) {
    long total = 0;
    int i = 1;
    while(true){
        total += (i * i);
        if(total>x){
            break;
        }
        i++;
    }
    return --i;
}

This can be also solved using transtion matrix and approach similar to fibonacci series etc. problems. Transition matrix:

(0 1 0 2 1 0) - i^2

(0 0 0 1 0 0) - i-1

(0 0 0 1 1 0) - i

(0 0 0 0 1 0) - const 1

(0 1 0 2 1 1) - Sum(1 to i inclusive)

Initial vector for i=1: (1, 0, 1, 1, 1)

Then use efficient exponentitation to find relevant matrices and multiple. O(log n)