vector<int> time(n);

这行代码定义了一个名为 `time` 的 `vector` 对象，其中 `vector<int>` 表示这是一个存储整数类型的 `vector` 对象。`n` 是一个变量或者常量，表示这个 `vector` 对象的长度，即它可以存储的整数个数。这个 `vector` 对象中的每一个元素都可以通过下标来访问和修改，例如 `time[0]` 表示访问第一个元素，`time[1]` 表示访问第二个元素，以此类推。在这个定义中，`time` 初始化为一个长度为 `n` 的空 `vector` 对象，即其中没有任何元素。如果需要给 `time` 中的元素赋值，可以使用如下方式：

time[0] = 1;   // 给第一个元素赋值为1
time[1] = 2;   // 给第二个元素赋值为2
// ...

也可以使用 `push_back` 函数向 `vector` 中添加元素，例如：

time.push_back(1);  // 向末尾添加一个元素1
time.push_back(2);  // 再向末尾添加一个元素2
// ...

这样，`vector` 对象中的元素个数会动态增加，不需要提前指定长度。

相关问题

Joseph question: n knight numbers 1, 2, n. Sitting around the round table. Knights numbered 1 start counting from 1, those who report to m are listed, and then the next position starts counting from 1 to find the last knight number left next to the round table. (1) Write a function template. Using a type of sequential container as a template parameter, solve the Joseph problem using a specified type of sequential container in the template. m. N is the formal parameter of the function template. (2) Call the function template with vector<int>, deque<int>, and list<int>as type parameters. When calling, set n to a larger number and m to a smaller number (such as n=100000, n=5). Observe the time it takes to call the function template in three scenarios. Note: To submit the answer to this question, only one sequential container type needs to be selected as the template parameter. [Input Form] Input for program reference (prompt text before numbers): Input n and m:7 3 Output Form Output of program reference: Result:4 Sample Input Input n and m:7 3 Sample output Result:4 Sample Description Scoring criteria #include<iostream> #include<vector> using namespace std; int main() { vector<int> a; int n,m,x=0; cout<<"Input n and m:"; cin>>n>>m; a.resize(n); for(int i=0;i<n;i++) { a[i]=i+1; } cout<<"Result:"<<a[0]<<endl; return 0; }

Here's the function template to solve the Joseph problem using a specified type of sequential container as the template parameter:

template<typename Container>
int joseph(int n, int m) {
    Container knights;
    for (int i = 1; i <= n; ++i) {
        knights.push_back(i);
    }
    auto it = knights.begin();
    while (knights.size() > 1) {
        for (int i = 1; i < m; ++i) {
            ++it;
            if (it == knights.end()) {
                it = knights.begin();
            }
        }
        it = knights.erase(it);
        if (it == knights.end()) {
            it = knights.begin();
        }
    }
    return knights.front();
}

The function takes two integer parameters `n` and `m`, which are the number of knights and the count to skip before removing a knight, respectively. The function returns the number of the last knight left standing.

To call the function with different sequential container types, we can use the following code:

int main() {
    int n = 100000;
    int m = 5;

    // Using vector<int>
    auto start_time = chrono::steady_clock::now();
    int result = joseph<vector<int>>(n, m);
    auto end_time = chrono::steady_clock::now();
    cout << "vector<int>: " << result << ", time: " << chrono::duration_cast<chrono::milliseconds>(end_time - start_time).count() << "ms" << endl;

    // Using deque<int>
    start_time = chrono::steady_clock::now();
    result = joseph<deque<int>>(n, m);
    end_time = chrono::steady_clock::now();
    cout << "deque<int>: " << result << ", time: " << chrono::duration_cast<chrono::milliseconds>(end_time - start_time).count() << "ms" << endl;

    // Using list<int>
    start_time = chrono::steady_clock::now();
    result = joseph<list<int>>(n, m);
    end_time = chrono::steady_clock::now();
    cout << "list<int>: " << result << ", time: " << chrono::duration_cast<chrono::milliseconds>(end_time - start_time).count() << "ms" << endl;

    return 0;
}

This code first sets `n` and `m` to the desired values. It then calls the `joseph` function with `vector<int>`, `deque<int>`, and `list<int>` as the sequential container types, and measures the time it takes to run each call using the `chrono` library. Finally, it prints the results and the execution times for each container type.

Note that the time it takes to run the function depends on the size of `n` and the value of `m`. For large values of `n` and small values of `m`, the function may take a long time to run, especially for `list<int>` which has slower insertions and deletions than `vector<int>` and `deque<int>`.

约瑟夫问题：n个骑士编号1，2，...，n，围坐在圆桌旁。编号为1的骑士从1开始报数，报到m的骑士出列，然后下一个位置再从1开始报数，找出最后留在圆桌旁的骑士编号。 （1）编写一个函数模板。以一种顺序容器的类型作为模板参数，在模板中使用指定类型的顺序容器求解约瑟夫问题。m,n是该函数模板的形参。 （2）分别以vector<int>,deque<int>,list<int>作为类型参数调用该函数模板，调用时将n设为较大的数，将m设为较小的数（例如令n=100000,n=5）。观察3种情况下调用该函数模板所需花费的时间。 注：本题答案的提交只需选择一种顺序容器类型作为模板参数。 【输入形式】 程序参考的输入（数字前为提示文字）： Input n and m:7 3 【输出形式】 程序参考的输出： Result:4 【样例输入】 Input n and m:7 3 【样例输出】 Result:4 【样例说明】 【评分标准】 #include<iostream> #include<vector> using namespace std; int main() { vector<int> a; int n,m,x=0; cout<<"Input n and m:"; cin>>n>>m; a.resize(n); for(int i=0;i<n;i++) { a[i]=i+1; } cout<<"Result:"<<a[0]<<endl; return 0; }

#include<iostream>
#include<vector>
#include<deque>
#include<list>
#include<chrono> // 用于计时
using namespace std;

// 定义函数模板
template<typename Container>
int josephus(int n, int m) {
    Container knights(n); // 创建容器并初始化
    for(int i = 0; i < n; i++) {
        knights[i] = i + 1;
    }
    int pos = 0; // 初始位置为0
    while(n > 1) {
        pos = (pos + m - 1) % n; // 计算出列位置
        knights.erase(knights.begin() + pos); // 移除骑士
        n--; // 骑士数减1
    }
    return knights[0]; // 返回最后剩余的骑士编号
}

int main() {
    int n, m;
    cout << "Input n and m: ";
    cin >> n >> m;

    // 使用vector容器
    auto start = chrono::steady_clock::now(); // 记录开始时间
    int result = josephus<vector<int>>(n, m);
    auto end = chrono::steady_clock::now(); // 记录结束时间
    cout << "Using vector: " << result << endl;
    cout << "Time cost: " << chrono::duration_cast<chrono::microseconds>(end - start).count() << " us" << endl;

    // 使用deque容器
    start = chrono::steady_clock::now();
    result = josephus<deque<int>>(n, m);
    end = chrono::steady_clock::now();
    cout << "Using deque: " << result << endl;
    cout << "Time cost: " << chrono::duration_cast<chrono::microseconds>(end - start).count() << " us" << endl;

    // 使用list容器
    start = chrono::steady_clock::now();
    result = josephus<list<int>>(n, m);
    end = chrono::steady_clock::now();
    cout << "Using list: " << result << endl;
    cout << "Time cost: " << chrono::duration_cast<chrono::microseconds>(end - start).count() << " us" << endl;

    return 0;
}