使用向量解决 Josephus 问题

Question

编辑：我似乎至少已经解决了错误，并更新了代码。然而，数学似乎仍然没有解决。有什么想法吗？

简而言之，我正在尝试用 C++ 编写一个程序，它会提示用户输入初始圈子中的人数，然后告诉他们如果 k（被处决前统计到的人数）= 3.

我有我认为正确的想法，但如果我输入 k 和其他任何东西一样，我会得到错误 "Debug Assertion Failed" 和 "Expression: vector erase iterator outside range"大于 1、2 或 5。

// ConsoleApplication2.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
#include <vector>
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
    int n;//size of the circle
    vector <int> circle; //the circle itself

    //ask for how many people are in the circle
    cin >> n;

    //fill the circle with 1,2,3...n
    for (int idx = 0; idx < n; idx++)
    {
        circle.push_back (idx+1);
    }


    //cout << "The size of the circle is " << circle.size() << ".\nThe highest number is " << circle[n-1] << "."; //test to make sure numbers are being assigned properly to each vector element


    for (int count = 0, idx = 0; circle.size() > 1; idx++,count++)
    {
        //if the position (idx) is greater than the size of the circle, go back to the beginning of the circle and start counting again
        if (idx >= circle.size())
        {
            idx = 0;
        }

        //every time the counter reaches three, that person is executed
        if (count == 3)
        {
            circle.erase (circle.begin()+(idx));
            count = 0;
        }
    }

    cout << "The place to stand to win the hand is position #" << circle.front() << ".\n";

    return 0;
}

Answer 1

您只需检查 if (idx > circle.size())，然后继续调用 circle.erase (circle.begin()+(idx));。 idx == circle.size().

时此调用不安全

Answer 2

@Pradhan 已经为您提供了原始错误的解决方案（idx >= circle.size() 而不是 idx >= circle.size()。

为什么结果仍然不正确：
当你擦除一个元素时，你必须调整你的索引来补偿它（减去 1）。否则索引对应于向量中的下一个元素，因此实际上你总是每 4 个人执行一次，而不是每 3 个人执行一次。

这是我的代码版本：

#include <iostream>
#include <vector>
#include <numeric>

using namespace std;

int main(){         
    int n;
    cin >> n;

    //fill the circle with 1,2,3...n
    vector <int> circle(n);
    std::iota(std::begin(circle), std::end(circle),1);

    int idx = -1;
    while (circle.size() > 1) {
        //advance by threee
        idx = (idx + 3) % circle.size();

        //execute Person 
        circle.erase(circle.begin() + (idx));

        //readjust to compensate for missing element
        --idx;
    }
    cout << "The place to stand to win the hand is position #" << circle.front() << ".\n";
}

当然可以把循环改写成

    int idx = 0;
    while (circle.size() > 1) {
        //advance by three
        idx = (idx + 2) % circle.size();

        //execute Person 
        circle.erase(circle.begin() + (idx));           
    }

使用向量解决 Josephus 问题

Solving the Josephus Problem using a vector

c++

vector

josephus