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Google codes are available...

Number of ways

Sequence gcd


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#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <numeric>

int calculate_gcd(int a, int b) {
if (a == 0) return b;
if (b == 0) return a;
return std::gcd(a, b);
}

long long dp[20][2][2][10][2];
bool visited[20][2][2][10][2];

std::string S;

long long solve(int idx, bool tight, bool is_leading_zero, int current_gcd_val, bool has_zero_digit) {
if (idx == S.length()) {
if (is_leading_zero) {
return 0;
}
if (has_zero_digit) {
return 0;
}
return current_gcd_val;
}

if (visited[idx][tight][is_leading_zero][current_gcd_val][has_zero_digit]) {
return dp[idx][tight][is_leading_zero][current_gcd_val][has_zero_digit];
}

long long ans = 0;
int upper_bound = tight ? (S[idx] - '0') : 9;

for (int digit = 0; digit <= upper_bound; ++digit) {
bool new_tight = tight && (digit == upper_bound);
bool new_is_leading_zero = is_leading_zero && (digit == 0);
bool new_has_zero_digit = has_zero_digit || (!is_leading_zero && digit == 0);

int next_gcd_for_param;
if (new_is_leading_zero) {
next_gcd_for_param = 0;
} else {
if (digit != 0) {
if (current_gcd_val == 0) {
next_gcd_for_param = digit;
} else {
next_gcd_for_param = calculate_gcd(current_gcd_val, digit);
}
} else {
next_gcd_for_param = current_gcd_val;
}
}

ans += solve(idx + 1, new_tight, new_is_leading_zero, next_gcd_for_param, new_has_zero_digit);
}

visited[idx][tight][is_leading_zero][current_gcd_val][has_zero_digit] = true;
return dp[idx][tight][is_leading_zero][current_gcd_val][has_zero_digit] = ans;
}

long long solve_up_to(long long N_val) {
if (N_val == 0) {
return 0;
}
S = std::to_string(N_val);

for (int i = 0; i < 20; ++i) {
for (int j = 0; j < 2; ++j) {
for (int k = 0; k < 2; ++k) {
for (int l = 0; l < 10; ++l) {
for (int m = 0; m < 2; ++m) {
visited[i][j][k][l][m] = false;
dp[i][j][k][l][m] = 0;
}
}
}
}
}
return solve(0, true, true, 0, false);
}

int main() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(NULL);

int T;
std::cin >> T;

while (T--) {
long long L, R;
std::cin >> L >> R;

long long ans_R = solve_up_to(R);
long long ans_L_minus_1 = solve_up_to(L - 1);

long long final_answer = ans_R - ans_L_minus_1;
std::cout << final_answer << std::endl;
}

return 0;
}

Sequence gcd βœ”οΈβœ… Google
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Infosys Previous Year Hiring Timeline (2024):

πŸ“’ Test Announced: 4th July 2024

πŸ“ Test Conducted: 7th July 2024


Based on my previous year’s experience, this timeline is highly likely to repeat this year as well. πŸ‘πŸ»βœοΈ

Stay prepared and alert for updates!

https://t.me/code_alphix/604
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