C. Bewitching Stargazer solution

[@computerknowledge4217]

C.
In C++ language
#include
using namespace std;

pair computeLucky(long long left, long long right, int threshold) {
if (right - left + 1 < threshold) {
return {0, 0};
}

if (right - left + 1 == 1) {
return {left, 1};
}

long long mid = left + (right - left) / 2;
if ((right - left + 1) % 2) {
pair leftResult = computeLucky(left, mid - 1, threshold);

long long totalLucky = mid + 2 * leftResult.first + mid * leftResult.second;
int totalSegments = 2 * leftResult.second + 1;
return {totalLucky, totalSegments};
} else {
pair leftResult = computeLucky(left, mid, threshold);

long long totalLucky = 2 * leftResult.first + mid * leftResult.second;
int totalSegments = 2 * leftResult.second;
return {totalLucky, totalSegments};
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
int test_cases;
cin >> test_cases;
while (test_cases--) {
long long n, k;
cin >> n >> k;

pair result = computeLucky(1, n, k);

cout

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