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编辑：我似乎至少已经解决了错误，并更新了代码。但是，数学似乎仍然没有解决。有任何想法吗？

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圆圈中的人数，然后告诉他们如果k（计算的人数在被执行之前）= 3。

我有我认为正确的想法，但是如果我将k输入为 1、2 或 5 以外的任何值，则会收到错误“调试断言失败”和“表达式：向量擦除迭代器超出范围”。

// 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;
}
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2 回答 2

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你只检查if (idx > circle.size())然后继续打电话circle.erase (circle.begin()+(idx));。当idx == circle.size().

于 2015-04-27T20:12:14.537 回答
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@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：
当你删除一个元素时，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，因此实际上您总是每 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，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));           
    }
于 2015-04-27T21:50:24.883 回答