knights={'a':'b','c':'d'} print

这段代码存在语法错误，应该在代码末尾加上括号来表示打印语句。修改后的代码如下：

knights = {'a': 'b', 'c': 'd'}
print(knights)

这段代码创建了一个名为knights的字典，其中包含两个键值对：{'a': 'b'}和{'c': 'd'}。然后使用print函数打印整个字典。输出结果为：

{'a': 'b', 'c': 'd'}

相关问题

knights = {'a': 'b', 'c': 'd'} for k,v in knights.items (): print(k,v)

这段代码存在语法错误，应该将for循环语句放在一个代码块中。同时，循环语句中的字典名称应该为knights而不是knights.items()。修改后的代码如下：

knights = {'a': 'b', 'c': 'd'}
for k, v in knights.items():
    print(k, v)

这段代码遍历了字典knights中的所有键值对，并使用print函数将每个键值对中的键和值打印出来。输出结果为：

a b
c d

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>`.