笔试面试相关记录（5）

（1）不包含重复字符的最长子串的长度

#include 
#include 
#include 

（2）小米基站问题，基站数组中，每个基站的x,y,q分别表示基站的坐标x和y，以及信号q，radius表示手机可以接收信号的范围，当手机距离基站的距离小于radius时，基站的信号为floor(q/(1+d))，求出一个信号最强的位置，如果有多个位置，按照字典序取第一个。

#include 
#include 
#include 
#include 

using namespace std;

double distance(int x1, int y1, int x2, int y2) {
    return sqrt(pow(x1-x2, 2) + pow(y1-y2, 2));
}

int main() {
    string str;
    std::getline(std::cin, str);
        int t = 0, n, radius;
        for (long unsigned i = 0; i < str.size(); i++) {
            if (isdigit(str[i])) {
                t = t*10 + (str[i]-'0');
            } else {
                n = t;
                t = 0;
            }
        }
        radius = t;
        int x, y, q;
        int min_x = 1e8, min_y = 1e8, max_x = 0, max_y = 0;
        vector> vec;
        for (int i = 0; i < n; i++) {
            scanf("%d,%d,%d", &x, &y, &q);
            vec.push_back({x,y,q});
            min_x = min(x, min_x);
            min_y = min(y, min_y);
            max_x = max(x, max_x);
            max_y = max(y, max_y);
        }
        vector> grid(max_x+1, vector(max_y+1, 0));
        int max_q = 0, ans_x = 1e8, ans_y = 1e8;
        for (int i = 0; i < n; i++) {
            int n_x = vec[i][0], n_y = vec[i][1], q = vec[i][2];
            for (int j = -radius; j <= radius; j++) {
                for (int k = -radius; k <= radius; k++) {
                    int new_x = n_x + j, new_y = n_y + k;
                    double dis = distance(new_x, new_y, n_x, n_y);
                    if (new_x >= 0 && new_y >= 0 && new_x <= max_x && new_y <= max_y && dis < radius) {
                        grid[new_x][new_y] +=  floor(q/(1+dis));
                        if (grid[new_x][new_y] > max_q) {
                            max_q = grid[new_x][new_y];
                            ans_x = new_x;
                            ans_y = new_y;
                        } else if (grid[new_x][new_y] == max_q && (new_x < ans_x || (new_x == ans_x && new_y < ans_y))) {
                            ans_x = new_x;
                            ans_y = new_y;
                        }
                    }
                }
            }
        }
        cout << ans_x << "," << ans_y << endl;
   
}

(3)是一个拓扑排序问题，

输入n，表示n个任务

输入1:0,0:1表示任务1依赖任务0，任务0依赖任务1,问这个任务能不能完成？可以完成输出1，否则输出0；

#include 
#include 
#include 
#include 
using namespace std;

int main() {
    int n;
    string str;
    while (cin >> n) {
        cin >> str;
        int t = 0;
        int a = 0, b = 0;
        int max_a = 0, max_b = 0;
        vector> vec;
        for (long unsigned i = 0; i < str.size(); i++) {
            if (isdigit(str[i])) {
                t = t*10 + (str[i]-'0');
            } else if (str[i] == ':') {
                a = t;
                t = 0;
            } else if (str[i] == ',') {
                max_a = max(max_a, a);
                max_b = max(max_b, t);
                vec.push_back({a, t});
                t = 0;
            }
        }
        max_a = max(max_a, a);
        max_b = max(max_b, t);
        vec.push_back({a, t});
        vector> grid(max_a+1, vector(max_b+1, 0));
        for (int i = 0; i < n; i++) {
            grid[vec[i][0]][vec[i][1]] = 1;
        }
        queue q;
        vector dege(n, 0);
        for (int i = 0; i < n; i++) {
            int count = 0;
            for (int j = 0; j < n; j++) {
                if (grid[i][j] = 1) {
                    count++;
                }
            }
            dege[i] = count;
            if (count == 0) {
                q.push(i);
            }
        }
        while (q.size()) {
            int node = q.front();
            q.pop();
            for (int i = 0; i < n; i++) {
                if (grid[i][node] == 1) {
                    grid[i][node] = 0;
                    dege[i]--;
                    if (dege[i] == 0) {
                        q.push(i);
                    }
                }
            }
        }
        bool flag = false;
        for (int i = 0; i < n; i++) {
            if (dege[i]) {
                flag = true;
                break;
            }
        }
        if (flag) {
            cout << 0 << endl;
        } else {
            cout << 1 << endl;
        }
        
    }
    return 0;
}

(4)一个数组，表示每个工人的工作能力，现有一个工作需要ceil(n/2.0)个工人去完成，并且要求这些工人的工作能力之和大于等于target，求有多少种安排工人的方法。

输入

1(测试组数）

5 10(n,target）

3 2 3 4 5

输出7

#include 
#include 
#include 
#include 
#include 
using namespace std;


void backtrace(vector& nums, int sum, long unsigned path, int& ans, long unsigned start, long unsigned worker_num, int target, string str, set& set_str) {
    if (path == worker_num && sum >= target && set_str.find(str) == set_str.end()) {
        ans++;
        set_str.insert(str);
        return;
    }
    if (start >= nums.size() || path > worker_num) return;
    int len = nums.size();
    // 这个语句放在for循环外面，用于记录上次的结果，不放入for循环中
    string t = str;
    for (long unsigned i = start; i < len; i++) {
        // 这里一个错误导致弄了半天也没有弄出来，
        if (path == 0) str = to_string(i);
        // 下面的代码之前写的是str = str + "-" + to_string(i);这样写是有错误的
        // 因为每次选或者不选，是从上一次的结果后面进行添加的，如果这里使用了str = str + "-" + to_string(i);
        // 就会导致str在for循环中会被修改。
        else str = t + "-" + to_string(i);
        // 如果当前位置选择，则在上一次的结果上加上这次的数字
        backtrace(nums, sum+nums[i], path+1, ans, i+1, worker_num, target, str, set_str);
        // 当前位置不选择
        backtrace(nums, sum, path, ans, i+1, worker_num, target, t, set_str);
    }
}

int main() {
    int T;
    while (cin >> T) {
        for (int times = 0; times < T; times++) {
            int n, target;
            cin >> n >> target;
            vector nums(n, 0);
            int t;
            for (int i = 0; i < n; i++) {
                cin >> t;
                nums[i] = t;
            }
            long unsigned path = 0;
            // 如果使用ceil函数，记得要将其变成float或者double形式，如果是整数，就会出错。
            long unsigned worker_num = (n%2==1)?(n+1)/2:(n/2);
            string str = "";
            int ans = 0;
            set set_str;
            backtrace(nums, 0, path, ans, 0, worker_num, target, str, set_str);          
            cout << ans << endl;
        }
    }
    return 0;
}

（5）

#include 
using namespace std; 
 
void func(int a, int b, int c) {
    c = a*b;
}

int main()
{
    int c;
    int a = 3, b = 3;
    func(a, b, c);
    cout << c << endl;
    printf("%d", c);
    return 0;
}
输出随机数，不是输出0

(6)

#include 
using namespace std; 
 
class T {
public:
    T() {cout << "T()" << endl;}
    
};
int main()
{
    T* p = new T[3];
    return 0;
}
输出
T()
T()
T()