Question

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

简而言之，我试图用C ++编写一个程序，它会提示用户输入初始圈子中的人数，然后告诉他们他们应该站在哪个位置以便在 k （执行前计算的人数）= 3.

我已经得到了我认为正确的想法，但我收到了错误＆＃34; Debug Assertion Failed＆＃34;和＆＃34;表达式：向量擦除迭代器超出范围＆＃34;如果我将 k 输入为1,2或5以外的任何其他内容。

// 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())，然后继续致电circle.erase (circle.begin()+(idx));。 idx == circle.size()时，此通话不安全。

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：
擦除元素时，必须调整索引以对其进行补偿（减去1）。否则索引对应于向量中的下一个元素，所以实际上你总是每隔4个人执行第3个人。

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，您可以将循环重写为

    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

2 个答案: