logoChat01.ai

Safe Salesforce PasswordsSalesforce provides a sec...

創建於：2026年1月23日

使用 GPT-5.2 Thinking 回答 Chat01

問題

Safe Salesforce PasswordsSalesforce provides a secure system for organizations to manage their accounts and data. To ensure maximum security, Salesforce has implemented password policies for user accounts. A password is considered safe if it satisfies the following criteria:The length of the password is exactly nn.The password consists of lowercase English characters only.The password does not contain kk or more consecutive identical characters.Given the integers nn (password length) and kk (maximum allowed consecutive identical characters), calculate the total number of distinct safe passwords that can be generated. Since the number can be large, return the result modulo (109+7)(10^9 + 7).ExampleConsider n=2,k=2n = 2, k = 2.The total number of passwords of length 2 is 26×26=67626 \times 26 = 676. There are 26 cases where k=2k = 2 consecutive characters are the same. Thus, the answer is 26×26−26=676−26=65026 \times 26 - 26 = 676 - 26 = 650, and 650 modulo (109+7)=650650 \text{ modulo } (10^9 + 7) = 650.Function DescriptionComplete the function countValidPasswords in the editor below.countValidPasswords has the following parameters:int n: the length of the passwordint k: the number of matching consecutive characters should be less than this numberReturnsint: the number of safe passwords, modulo (109+7)(10^9 + 7)Constraints2 \le n \le 10^5$$2 \le k \le nCode Editor (C++14)C++#include <bits/stdc++.h>

/*

  • Complete the 'countValidPasswords' function below.
  • The function is expected to return an INTEGER.
  • The function accepts following parameters:
    1. INTEGER n
    1. INTEGER k
      */

int countValidPasswords(int n, int k) {

}

int main() ...give me cpp code

思考

Planning efficient valid password count via dynamic programming

I need to calculate how many valid passwords of length n can be formed using 26 letters, without any run of identical letters exceeding length k-1. Using dynamic programming, I'll model valid strings with limited run lengths, applying a recurrence and possibly inclusion-exclusion to handle constraints efficiently.

Optimizing run length tracking for valid passwords

I need to track run lengths and use dynamic programming to efficiently calculate valid strings. The new approach uses recurrences to calculate sums without O(k) complexity each step. Essentially, I’m optimizing the process by maintaining a total sum and tracking forbidden run extensions.

Verifying recurrence and special cases for counting

I need to refine the recurrence by verifying how a[i][k-1] should be computed for cases where i >= k-1. For smaller cases, like when k=2, the recurrence behaves differently, especially for the base case. I'll need to include special handling for cases like a[1][1] where things initially deviate.

Refining recurrence calculation for valid password length

I’m defining base cases and refining the recurrence for calculating valid passwords. For lengths less than k, no run of length k-1 occurs, so I use S_i = 26^i. When i equals k, I need to subtract strings with a run of k at the end. I’ll test with a smaller example to verify.

Refining recurrence for valid strings with max run length k=2

I’m reworking the recurrence formula for k=2. My earlier formula didn't match the expected result, so I adjusted it. For k=2, I need to handle the case where the formula for a[i][k-1] doesn’t apply, as the transition needs to work differently for this case. I’ll derive a more general formula that works consistently for all k.

Refining recurrence formula and validating cases

I'm refining the recurrence formula for situations where the maximum run length is constrained. The formula seems to hold: for larger values of i, I subtract terms based on previous states, and testing for specific cases like k=2 and n=3 confirms that the recurrence works. The formula S_i = 26*S_{i-1} - 25*S_{i-k} behaves as expected.

Finalizing password counting algorithm in C++

I'm finalizing the recurrence for counting valid passwords. For i>k, the recurrence formula subtracts 25*dp[i-k], and I’m ensuring everything works as expected. Now I need to compute the powers of 26 modulo 1e9+7 using efficient multiplication. The function structure is clear, and I’ll provide the full C++ code including input

Thought for 2m 4s

回答

cpp
#include <bits/stdc++.h>
using namespace std;

static const long long MOD = 1000000007LL;

// Count length-n strings over 26 lowercase letters with no run of identical chars of length >= k
int countValidPasswords(int n, int k) {
    vector<long long> dp(n + 1, 0);
    dp[0] = 1; // empty

    // For i < k, constraint is inactive => 26^i
    long long p26 = 1;
    int upto = min(n, k - 1);
    for (int i = 1; i <= upto; i++) {
        p26 = (p26 * 26LL) % MOD;
        dp[i] = p26;
    }

    if (n < k) return (int)dp[n];

    // i == k: all strings minus the 26 strings of all-same letter
    dp[k] = (26LL * dp[k - 1] - 26LL) % MOD;
    if (dp[k] < 0) dp[k] += MOD;

    // i > k: dp[i] = 26*dp[i-1] - 25*dp[i-k]
    for (int i = k + 1; i <= n; i++) {
        dp[i] = (26LL * dp[i - 1] - 25LL * dp[i - k]) % MOD;
        if (dp[i] < 0) dp[i] += MOD;
    }

    return (int)dp[n];
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, k;
    cin >> n >> k;
    cout << countValidPasswords(n, k) << "\n";
    return 0;
}

問題

xSalesforce Data CollectionA Salesforce administrator is working on optimizing data collection for Opportunity records across multiple branches.The branches follow a hierarchical structure, forming a tree with nn nodes labeled from 0 to n−1n-1. Each node represents a branch, and an array opportunityData of size nn indicates whether a branch contains important Opportunity data. Specifically, opportunityData[i] is 1 if branch ii holds relevant data and 0 otherwise.To gather all Opportunity data, the admin must navigate through the branches using bidirectional paths (edges), which can be traversed multiple times if necessary. The goal is to minimize the total distance traveled, measured by the number of edges crossed. Additionally, any data within a radius of 2 edges from a branch can be collected. For system consistency, the retrieval process must start and end at the same branch.Determine the minimum number of edges the admin needs to traverse to collect all Opportunity data.Function DescriptionComplete the function minEdgesToCollectData in the editor.The function has the following parameters:vector<int> opportunityData: An array indicating whether each branch contains Opportunity data.int branch_nodes: The number of branches in the hierarchy.vector<int> branch_from: The starting point of each connection.vector<int> branch_to: The ending point of each connection.Returns:int: The minimum number of edges required to collect all data.Constraints2 \le \text{branch\_nodes} \le 10^5$$\text{opportunityData}[i] = 0 or 11, where 0 \le i < \text{branch\_nodes}$$0 \le \text{branch\_from}[i] < \text{branch\_nodes} where 0 \le i < \text{branch\_nodes} - 1$$0 \le \text{branch\_to}[i] < \text{branch\_nodes} where 0≤i<branch_nodes−10 \le i < \text{branch\_nodes} - 1Sample Case 1Sample Input:n = branch_nodes = 10opportunityData = [1, 0, 1, 1, 1, 1, 0, 0, 1, 1]Edges: (0,1), (0,2), (0,3), (0,4), (0,5), (1,6), (6,7), (7,8), (5,9)Sample Output:4Explanation:All Opportunity data, except the data at Branch 8, is collected from Branch 0. The admin follows the path 0→1→6→1→00 \to 1 \to 6 \to 1 \to 0, then collects the data at Branch 8 from Branch 6.int minEdgesToCollectData(vector<int> opportunityData, int branch_nodes, vector<int> branch_from, vector<int> branch_to) {
// Your code here
}give me cpp code

思考

Planning minimum edge traversal for coin collection

I need to write a C++ code to solve for the minimum edge traversals to collect all marked nodes with a radius 2 collection, where the start and end are the same. This problem aligns closely with LeetCode 2603, "Collect Coins in a Tree", where we traverse edges after pruning leaves without coins. The key is figuring out the minimum path to collect everything, considering the radius condition.

Pruning and traversal strategy for coin collection

The solution involves pruning leaves with no coins, followed by two rounds of trimming all leaves, regardless of coin status. After this, we compute the remaining edges and double their count for the round trip. The key steps in the implementation are building the adjacency list, processing leaves, and performing two pruning rounds.

Process for pruning and counting edges

For two rounds: I process all nodes in the queue, remove leaves, decrease the degree, and push neighbors that become leaves into the next queue. After trimming, I count the remaining edges by summing the degrees of nodes and dividing by 2. Finally, I return the remaining edges multiplied by 2.

Finalizing pruning process and counting edges

Leaf node 6 isn't trimmed in phase 1. After phase 2, I remove leaves 2, 3, 4, 9, and 8, then continue reducing degrees. After trimming, the count of remaining edges is 2, resulting in the final answer of 4. I'll finalize the function signature without needing mod or additional complexity.

Thought for 42s

回答

cpp
#include <bits/stdc++.h>
using namespace std;

int minEdgesToCollectData(vector<int> opportunityData,
                          int branch_nodes,
                          vector<int> branch_from,
                          vector<int> branch_to) {
    int n = branch_nodes;
    vector<vector<int>> adj(n);
    vector<int> deg(n, 0);

    for (int i = 0; i < n - 1; i++) {
        int u = branch_from[i], v = branch_to[i];
        adj[u].push_back(v);
        adj[v].push_back(u);
        deg[u]++; deg[v]++;
    }

    vector<char> removed(n, 0);
    queue<int> q;

    // 1) Prune all leaves with no data (opportunityData[u] == 0)
    for (int i = 0; i < n; i++) {
        if (deg[i] == 1 && opportunityData[i] == 0) q.push(i);
    }
    while (!q.empty()) {
        int u = q.front(); q.pop();
        if (removed[u]) continue;
        removed[u] = 1;
        for (int v : adj[u]) {
            if (removed[v]) continue;
            if (--deg[v] == 1 && opportunityData[v] == 0) q.push(v);
        }
        deg[u] = 0;
    }

    // 2) Remove two layers of remaining leaves (because you can collect within distance 2)
    for (int i = 0; i < n; i++) {
        if (!removed[i] && deg[i] == 1) q.push(i);
    }

    for (int layer = 0; layer < 2; layer++) {
        int sz = (int)q.size();
        while (sz--) {
            int u = q.front(); q.pop();
            if (removed[u]) continue;
            removed[u] = 1;
            for (int v : adj[u]) {
                if (removed[v]) continue;
                if (--deg[v] == 1) q.push(v);
            }
            deg[u] = 0;
        }
    }

    // 3) Remaining edges must be traversed twice (go and return)
    long long remainingEdgeCountTimes2 = 0;
    for (int i = 0; i < n; i++) {
        if (!removed[i]) remainingEdgeCountTimes2 += deg[i];
    }
    long long remainingEdges = remainingEdgeCountTimes2 / 2;
    return (int)(2 * remainingEdges);
}

分享這個問答