The code editor gives you code for the following problem statement. However, the solution fails the test cases because of code bugs. Your task is to find and fix all the bugs so that it passes all the test cases.
You are given two arrays, A and B, of size N consisting of non-negative integers. You have to select K numbers from B and take the bitwise OR of all the array elements in A with the chosen K numbers and add their sum. In other words, take each element from array A and add its bitwise OR with each chosen element from array B to the sum.
Note: You cannot choose two adjacent elements in B.
You have to find the minimum value of K such that the sum is greater than or equal to M. Print -1 if impossible.
K = 0, then sum = 0.K = 1, take (2) as the chosen element from B, then:sum = (1|2) + (2|2) + (3|2) + (4|2) + (5|2) = 3 + 2 + 3 + 6 + 7 = 21K = 2, take (2, 2) as the chosen elements from B, then:sum= (1|2) + (1|2) + (2|2) + (2|2) + (3|2) + (3|2) + (4|2) + (4|2) + (5|2) + (5|2) = (3 + 2 + 3 + 6 + 7) * 2 = 42K = 2 is the minimum required to make sum >= 40.Complete the MinReq function provided in the editor. This function takes the following 4 parameters and returns the required answer:
K elements.Note: This is the input format that you must use to provide custom input (available above the Compile and Test button).
T which denotes the number of test cases.T cases are defined as follows:N, denoting the size of the array.N space-separated integers denoting the array A.N space-separated integers denoting the array B.M, the lower limit for the sum.Print an integer value that is the minimum K for each of the test cases. Each value that is printed must be line-separated.
1 <= T <= 10
1 <= N <= 10^3
0 <= A, B <= 10^9
0 <= M <= 10^18
import java.util.Scanner;
public class Main {
static int MinReq(int N, int[] A, int[] B, long M) {
long[][] dp = new long[N + 1][N + 1]; // DP array to store max sums
long[] val = new long[N]; // Array to store OR values
// Calculate the OR values for each element in A with every element in B
for (int j = 0; j < N; j++) {
val[j] = 0; // Initialize val[j]
for (int i = 0; i < N; i++) {
val[j] += (A[i] | B[j]); // Calculate A[i] OR B[j]
}
}
// Initialize DP table
for (int i = 0; i <= N; i++) {
for (int j = 0; j <= N; j++) {
dp[i][j] = Long.MIN_VALUE; // Initialize with minimum value
}
}
dp[0][0] = 0; // Base case: using 0 elements gives a sum of 0
// Fill the DP table
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= N; j++) {
dp[i][j] = dp[i][j - 1]; // Not choosing current B[j]
// Choose current B[j], ensure we do not pick adjacent elements
if (j >= 2) {
dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - 2] + val[j - 1]); // Choosing B[j-1]
}
}
}
// Find the minimum K such that the sum is >= M
for (int k = 1; k <= N; k++) {
for (int j = 0; j <= N; j++) {
if (dp[k][j] >= M) {
return k; // Minimum K found
}
}
}
return -1; // If not possible to reach M
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
// Read number of test cases
int T = sc.nextInt();
// Loop through each test case
for (int t = 0; t < T; t++) {
// Read the size of the arrays
int N = sc.nextInt();
// Read array A
int[] A = new int[N];
for (int i = 0; i < N; i++) {
A[i] = sc.nextInt();
}
// Read array B
int[] B = new int[N];
for (int i = 0; i < N; i++) {
B[i] = sc.nextInt();
}
// Read the lower limit for the sum M
long M = sc.nextLong();
// Call the MinReq function and print the result
int result = MinReq(N, A, B, M);
System.out.println(result);
}
sc.close();
}
}
#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>
using namespace std;
int MinReq(int N, vector<int>& A, vector<int>& B, long long M) {
vector<vector<long long>> dp(N + 1, vector<long long>(N + 1, LLONG_MIN)); // DP array to store max sums
vector<long long> val(N); // Array to store OR values
// Calculate the OR values for each element in A with every element in B
for (int j = 0; j < N; j++) {
val[j] = 0; // Initialize val[j]
for (int i = 0; i < N; i++) {
val[j] += (A[i] | B[j]); // Calculate A[i] OR B[j]
}
}
dp[0][0] = 0; // Base case: using 0 elements gives a sum of 0
// Fill the DP table
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= N; j++) {
dp[i][j] = dp[i][j - 1]; // Not choosing current B[j-1]
// Choose current B[j-1], ensure we do not pick adjacent elements
if (j >= 2) {
dp[i][j] = max(dp[i][j], dp[i - 1][j - 2] + val[j - 1]); // Choosing B[j-1]
}
}
}
// Find the minimum K such that the sum is >= M
for (int k = 1; k <= N; k++) {
for (int j = 1; j <= N; j++) {
if (dp[k][j] >= M) {
return k; // Minimum K found
}
}
}
return -1; // If not possible to reach M
}
int main() {
int T;
cin >> T;
// Loop through each test case
for (int t = 0; t < T; t++) {
// Read the size of the arrays
int N;
cin >> N;
// Read array A
vector<int> A(N);
for (int i = 0; i < N; i++) {
cin >> A[i];
}
// Read array B
vector<int> B(N);
for (int i = 0; i < N; i++) {
cin >> B[i];
}
// Read the lower limit for the sum M
long long M;
cin >> M;
// Call the MinReq function and print the result
int result = MinReq(N, A, B, M);
cout << result << endl;
}
return 0;
}
A function S(x) is defined as the sum of all divisors of x. For example:
S(4) = 1 + 2 + 4 = 7S(10) = 1 + 2 + 5 + 10 = 18Now, let us define a function F(x) as the sum of S(y) where y is a divisor of x, i.e., y divides x.
You are given multiple tasks. In each task, you are provided two integers L and R. Your task is to calculate the sum of F(x) for all x from L to R, i.e., compute the sum F(L) + F(L+1) + ... + F(R).
Q denoting the number of tasks.Q lines each contain two space-separated integers L and R representing the lower and upper bounds of a range.For each task, print a single integer in a new line representing the sum of F(x) for all x in the range [L, R].
1 ≤ Q ≤ 100,0001 ≤ L ≤ R ≤ 1,000,000For the given example where L = 8 and R = 10:
F(8) = S(1) + S(2) + S(4) + S(8) = 1 + 3 + 7 + 15 = 26F(9) = S(1) + S(3) + S(9) = 1 + 4 + 13 = 18F(10) = S(1) + S(2) + S(5) + S(10) = 1 + 3 + 6 + 18 = 28Thus, F(8) + F(9) + F(10) = 26 + 18 + 28 = 72.
#include <bitset>
#include <algorithm>
#include <cassert>
#include <cctype>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <limits>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
#include <fstream>
#include <sstream>
using namespace std;
vector<int> S(1000001, 0);
vector<int> F(1000001, 0);
vector<long long> prefix_sum(1000001, 0);
signed main()
{
for (int i = 1; i <= 1e6; i++)
{
for (int j = i; j <= 1e6; j += i)
{
S[j] += i;
}
}
for (int i = 1; i <= 1e6; i++)
{
for (int j = i; j <= 1e6; j += i)
{
F[j] += S[i];
}
}
for (int i = 1; i <= 1e6; i++)
{
prefix_sum[i] = prefix_sum[i - 1] + F[i];
}
int n;
cin >> n;
while (n--)
{
int a, b;
cin >> a >> b;
cout << prefix_sum[b] - prefix_sum[a - 1] << endl;
}
return 0;
}
Given a sequence of n integers a1, a2, ..., an, you are asked to perform the following operation and return the obtained result:
For a given integer x, calculate the value of the expression min(m₁, m₂, ..., mz-1) where:
fun(a_i, a_{i+1}, ..., a_z-1) returns the value of the rightmost number with the highest number of distinct prime factors.m_i = fun(a_i, a_{i+1}, ..., a_z-1)In other words, you need to find the minimum of the values m₁, m₂, ..., m_{z-1}.
Complete the solve function provided in the editor. The function takes the following 3 parameters and returns the answer:
x: Represents the given integer value.n: Represents the size of the array a.a: Represents the values in the array.The first line contains two integers x and n.
The second line contains n space-separated integers representing the array a.
Print an integer denoting the value of the expression.
1 ≤ n ≤ 100 ≤ a_i ≤ 10^6m₁ = fun(a₁, a₂, a₃) = 6m₂ = fun(a₂, a₃, a₄) = 10m₃ = fun(a₃, a₄, a₅) = 10Thus, min(m₁, m₂, m₃) = 6.
Your code must print the correct output for the provided sample input. However, it will be tested against hidden test cases, so make sure your solution is optimized and correct.
Given a tree with N vertices labeled from 1 to N, rooted at vertex 1. The tree is an undirected graph with N-1 edges. Each edge i connects two vertices a_i and b_i. Each vertex i is colored either white (represented by '0') or black (represented by '1').
The beauty of a vertex i is defined as the number of paths in its subtree that have end vertices of the opposite color.
You are required to find the beauty of all N vertices.
i is a connected subgraph consisting of all the descendants of i, including i.Complete the function solve. This function takes the following 3 parameters:
N: Represents the number of vertices in the tree.Color: A string representing the color of the vertices ('0' for white and '1' for black).Edges: A list of N-1 pairs, each denoting an undirected edge connecting two vertices.The function should return the beauty for all N vertices as space-separated integers.
N, denoting the number of vertices in the tree.Color of size N, consisting of characters '0' (white) and '1' (black).N-1 lines contain two integers a_i and b_i, denoting an undirected edge between the vertices numbered a_i and b_i.Print N space-separated integers, where each integer denotes the beauty of the corresponding vertex.
1 ≤ N ≤ 10^5Color[i] ∈ {0, 1} for all i in [1, N]1 ≤ a_i, b_i ≤ N for all i in [1, N-1]5
01101
3 1
4 3
5 3
2 4
4 0 3 0 0
For the given tree:
Thus, the answer is 4 0 3 0 0.
You bought two virtual reality glasses. Each glasses has a unique feature:
The maze is represented as a grid of size N x M. You start at the top-left corner of the grid and your goal is to reach the bottom-right corner. You can move vertically or horizontally, and each movement takes 1 minute.
Some cells in the maze contain obstacles:
'#': Represents a door, passable only with the first glasses.'+': Represents a wall, passable only with the second glasses.'.': Represents a clear passage, which can be traversed by both glasses.You need to determine how long it takes to reach the bottom-right corner of the maze using each pair of glasses. If it is impossible to reach the goal with either pair, return -1 for that pair.
Complete the function solve. This function takes the following 3 parameters and returns an array of length 2:
N: An integer representing the number of rows in the grid.M: An integer representing the number of columns in the grid.maze: A list of N strings representing the maze, where each character in maze[i] can be:
. for a clear passage,# for a door,+ for a wall.The function should return an array with two integers:
If it's impossible to reach the bottom-right corner using either glasses, return -1 for that glasses.
N and M, denoting the number of rows and columns in the grid.N lines contain strings of length M, where each string represents a row of the maze.If reaching the bottom-right corner is not possible with either glasses, print -1.
1 ≤ N, M ≤ 500. for paths,# for doors,+ for walls.5 5
..#..
.#.#.
..#..
#+.#.
…..
-1
8
-1.You are given an array of integers of size N. You need to find the minimum possible sum that can be formed by adding all the elements in the array after performing the following operation at most once:
i. If the sum of all the elements from index 0 to index i-1 is divisible by the element at index i, divide the sum by the element at index i.i to the resultant sum. Instead, you will add all other elements from index i+1 to the end of the array.Return the minimum possible sum that can be obtained after performing the operation once.
i is chosen, the sum of elements from index 0 to i-1 (denoted as S) must be divisible by the element at index i.i in the final sum, but include all the elements after index i.Complete the function solve. This function takes the following 2 parameters and returns a 64-bit integer representing the minimum possible sum:
N: An integer representing the size of the array.Arr: An array of size N containing integers.T.N, the size of the array.N integers, representing the elements of the array.For each test case, output a single integer representing the minimum possible sum after performing the operation.
1 ≤ T ≤ 5001 ≤ N ≤ 101 ≤ Arr[i] ≤ 10N across all test cases will not exceed 10^5.2
5
1 4 5 5 10
3
100 2 1
12
3
The first line contains T, denoting the number of test cases.
For the first test case:
Given:
N = 5Arr = [1, 4, 5, 5, 10]We evaluate the possible operations for each index:
0. Dividing 0 by Arr[0] = 1, we get: - 0 / 1 = 0. The sum after the operation is 0 + 4 + 5 + 5 + 10 = 24.1 + 4 = 5. Dividing 5 by Arr[2] = 5, we get: - (1 + 4) / 5 + 5 + 10 = 16.1 + 4 + 5 = 10. Dividing 10 by Arr[3] = 5, we get: - (1 + 4 + 5) / 5 + 10 = 12.1 + 4 + 5 + 5 = 15. Dividing 15 by Arr[4] = 10, we get: - 1 + 4 + 5 + 5 + 10 = 25.
Out of all possible places to apply the operation, the minimum possible result is 12.
Here’s the provided problem statement formatted in Markdown:Description:
Given an integer array nums of size N and an integer K, find the largest positive number less than or equal to (10^9) such that at least K numbers from the array nums are divisible by it.
- The first line contains an integer `T`, representing the number of test cases.
- For each test case:
- The first line contains an integer `N`, denoting the size of array `nums`.
- The second line contains integer `K`.
- The third line contains `N` spaced integers of the array `nums`.
- For each test case, print the required answer on a new line.
1<= T <= 1000
1<= N <= 10^5
1<= K <= N
1<= nums[i] <= 10^6, where 1<= i <= N
2
5
3
12 16 32 2 45
5
3
12 16 32 2 0
4
16
- In the first test case, the largest positive integer that divides at least 3 numbers in the array is 4.
- In the second test case, the largest positive integer that divides at least 3 numbers in the array is 16.