特殊的约瑟夫问题

版权声明：吸吸 https://blog.csdn.net/walk_dog/article/details/79313736

先看一下规矩的链表做法吧

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

int n, m, b;
struct People {
	int Num;
	People *Next;
};
People *Head, *Temp;

void Cycle()
{
	int i = 0;
	Head = new People;
	Temp = Head;
	while(i < n)
	{
		i++;
		Temp->Num = i;
		if(i == n)
		{
			Temp->Next = Head;
		} 
		else Temp->Next = new People;
		Temp = Temp->Next;
	}
	return;
}

int main()
{
	while(cin>>n>>m && n && m)
	{
		Cycle();
		People *Now;
		Now = Head;
		while(Now->Next != Now)
		{
			int x = 0;
			while(x < m-1)
			{
				x++;
				Now = Now->Next;
			}
			Now->Next = Now->Next->Next;
		}
		Now->Num--;
		if(Now->Num == 0) cout<< n << endl;
		else cout << Now->Num << endl;
	}
	return 0;
}

附代码↓ ↓ ↓

注释都在程序里：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <stack>
#include <queue> 

long long n, m;//说好的数据很水，但还是需要开到longlong型，有点难过 

void run()
{
    scanf("%lld %lld", &n, &m);//输入n和m，n个人，数到m的人出列 
    if (m == 2)//一种特殊情况需要自（chao）己（lai）的推理 
    {
        long long t = 1;
        while ((t << 1) <= n) t <<= 1;//2的r次方加上s，使s尽量小 
        printf("%lld",(n - t) * 2 + 1);//就输出 
        return;//返回 
    }
    else
    {
        int f = 0;
        for (int i = 2; i <= n; i++)
        {
            int g = f;
            f = (g + m) % i;//一个从网上找来的公式..自己下来还需要推一下 
        }
        printf("%d", f + 1);
    } 

}

int main()
{
//#ifndef LOCAL ←托哥上课时讲的新操作，有点骚 
//	freopen("joseph.in", "r", stdin);
//	freopen("joseph.out", "w", stdout);
//#endif
    run();//调用函数 
    return 0;
}