约瑟环问题

/*
约瑟环问题其实就是一个循环链表的问题 
*/ 
#include <iostream>
#include<ctime>
using namespace std;

//动态规划的思想，有点类似解决主元素问题的思路 
//巧妙的用0，1数组代替链表节点的删除 
int sun1(const int N,const int C)
{
    int i=0,j=0,n=N,s=0;
    int t=0;
    int *a=new int [N];
    for (i=0;i<N;i++)
    {
        a[i]=1;
    }
    
    while(0!=n)
    {
        s+=a[j]; //前面的数相加每加够C出列一个数 
        if(C==s)
        {
            a[j]=0;
            s=0; //要清0 
            --n;
            if(0!=n)
                cout<<j+1<<"->"; //s[j]实际对应的是第j+1个人 
            else
            {
                cout<<j+1<<endl;
                t=j+1;
            }
                
        }
        //j=(j+1)%N; // 循环队列，这里处理很巧妙
        j= (j==N-1)?0:j+1;
    }
    delete []a;
    return t;
    
}
/*
递推的方法 
N=1,winner=0;
N=i,winner=(winner(0)+C)%i;
winner +=1;
*/
int sun2(const int N, const int C)
{
    int f=0;
    for(int i=2;i<=N;i++)
    {
        f=(f+C)%i;
    }
    return f+1;
}
/*循环链表 */ 
int sun3(const int N,const int C)
{
    struct node
    {
        int num;
        node *next;
        node(const int i)
        {
            num=i;
            next=NULL;
        }
    };
    //初始化链表 
    node *first=NULL;
    node *opp=first;
    for(int i=1;i<=N;i++)
    {
        node *t=new node(i);
        if(first==NULL)
            first=t;
        else
            opp->next=t;
        opp=t;
    }
    
    opp->next=first;
    int winner=0;
    opp=first;
    while(true)
    {    
        for(int i=2;i<=C-1;i++)// 这里注意找到的是出列的节点的父节点 
            opp=opp->next;
        node *temp=opp->next;
        cout<<temp->num;
        if(opp!=temp)
        {
            cout<<"->";
            //删除节点 
            opp->next=temp->next;
            opp=opp->next;
            delete temp;
        }
        else
        {
            winner=temp->num;
            cout<<"
";    
            delete opp;
            break;
        }
    }
    return winner;
        
}
    
    

int main()
{
    ios::sync_with_stdio(false);
    int N=0,C=0;
    cout<<"Please enter the number of people:N=";
    cin>>N;
    cout<<"Please enter:C=";
    cin>>C;
    cout<<"winner is "<<sun1(N,C)<<endl;
    clock_t end1=clock();
    cout<<"winner is "<<sun2(N,C)<<endl;
    clock_t end2=clock();
    cout<<(end2-end1)/float(CLOCKS_PER_SEC)<<"seconds"<<endl;
    cout<<"winner is "<<sun3(N,C)<<endl;
    return 0;
}