科软25机试

题目:

2025科软复试上机题（回忆版）题解_哔哩哔哩_bilibili

1. 字符串反转

#include<bits/stdc++.h>
using namespace std;void solve(string& a, int CurN) {if (!(CurN % 2)) {int right = a.size() - 1;int left = 0;while (left < right) {swap(a[left], a[right]);left++;right--;}}
}void print(vector<string>& ans) {for (int i = 0; i < ans.size(); i++) {cout << ans[i] << " ";}cout << endl; 
}int main() {string s;getline(cin, s);stringstream ss(s);string tmpStr;int count = 0;vector<string> ans;while (getline(ss, tmpStr, ' ')) {ans.push_back(tmpStr);}for (int i = 0; i < ans.size(); i++) {solve(ans[i], i + 1);}print(ans);return 0;
}

2.进制转换

思路：将32进制字符串转换为十进制数;将十进制数转换为八进制字符串
#include<bits/stdc++.h>
using namespace std;
unordered_map<char, int> base32ToDec = {{'0', 0}, {'1', 1}, {'2', 2}, {'3', 3}, {'4', 4}, {'5', 5}, {'6', 6}, {'7', 7},{'8', 8}, {'9', 9}, {'A', 10}, {'B', 11}, {'C', 12}, {'D', 13}, {'E', 14}, {'F', 15},{'G', 16}, {'H', 17}, {'I', 18}, {'J', 19}, {'K', 20}, {'L', 21}, {'M', 22}, {'N', 23},{'O', 24}, {'P', 25}, {'Q', 26}, {'R', 27}, {'S', 28}, {'T', 29}, {'U', 30}, {'V', 31}
};
int solve1(string& Str32) {int decimalValue = 0;int length = Str32.length();for (int i = 0; i < length; ++i) {char ch = Str32[i];decimalValue += base32ToDec[ch] * static_cast<int>(pow(32, length - i - 1));}return decimalValue;
}
string solve2(int decimalValue) {string octalStr;while (decimalValue > 0) {octalStr += to_string(decimalValue % 8);decimalValue /= 8;}reverse(octalStr.begin(),octalStr.end());return octalStr.empty() ? "0" : octalStr;
}
int main() {string Str32; getline(cin,Str32);int decimalValue = solve1(Str32);string octalStr = solve2(decimalValue);cout<<octalStr<<endl;return 0;
}

3. 区间移除

#include<bits/stdc++.h>>
using namespace std;
int solve(vector<pair<int, int>>& arr) {if (arr.empty()) return 0;sort(arr.begin(), arr.end(), [](pair<int, int> a, pair<int, int> b) {return a.second < b.second; });int count = 0; int end = INT_MIN;for (auto& it :arr) {if (it.first >= end) {end = it.second;} else {count++;}}return count;
}
int main() {int n;cin >> n;vector<pair<int, int>> ans(n);for (int i = 0; i < n; i++) {cin >> ans[i].first >> ans[i].second;}cout << solve(ans) << endl;return 0;
}

4. 机器人

#include<bits/stdc++.h>
using namespace std;
int m, n;
vector<vector<int>> grid;
vector<vector<bool>> visited;
int maxValue = -1;
int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};
void dfs(int x, int y, int sum) {if (x == m - 1 && y == n - 1) {maxValue = max(maxValue, sum);return;}visited[x][y] = true;for (int i = 0; i < 4; i++) {int nx = x + dx[i];int ny = y + dy[i];if (nx >= 0 && nx < m && ny >= 0 && ny < n && !visited[nx][ny] && grid[nx][ny] != -1) {dfs(nx, ny, sum + grid[nx][ny]);}}visited[x][y] = false;
}
int main() {cin >> m >> n;grid.resize(m, vector<int>(n));visited.resize(m, vector<bool>(n, false));for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {string value;cin >> value;if (value == "#") {grid[i][j] = -1; } else {grid[i][j] = stoi(value);}}}if (grid[0][0] == -1 || grid[m-1][n-1] == -1) {cout << -1 << endl;return 0;}dfs(0, 0, grid[0][0]);cout << maxValue << endl;return 0;
}