何书帆月吗

b=input().split()
n=int(b[0])
m=int(b[1])
a=[]
k=-1
for i in range(1,n+1):
    a.append(i)
while len(a)!=1:
    k+=m
    k=k%len(a)
    print(a[k],end=' ')
    a.pop(k)
    k-=1
    
print(end='\n')
print(a[0])

最短好吧

单向链表（循环链表，即尾节点指针指向头结点）

class Node:
    def __init__(self,val:int,nxt:int):
        self.val=val
        self.nxt=nxt
def add (it,val):
    a.append(Node(val,a[it].nxt))
    a[it].nxt=len(a)-1
def delete (it):
    a[it].nxt=a[a[it].nxt].nxt
n,m=map(int,input().split())
a=[Node(i+1,i+1) for i in range(n)]
a[n-1].nxt=0
it=0
f=1
while n>1:
    n-=1
    for i in range(m-1-f):
        it=a[it].nxt
    print(a[a[it].nxt].val,end=" ")
    delete(it)
    f=0
print()
it=a[it].nxt
print(a[it].val)

n,m=map(int,input().split())
cnt=0
h=0
t=n
a=[i for i in range(1,n+1)]
cnt=0
while h<t-1:
    for i in range(m-1):
        a.append(a[h])
        h+=1
        t+=1
    print(a[h],end=" ")
    h+=1
print()
print(a[h])

python队列 h:队头（head） t:队尾（tail）

python

N=1e5+10
a=[0]*110000
p=input().split()
n=int(p[0])
m=int(p[1])
c=0
k=0
now=0
while(c!=(n-1)):
    now=(now+1)%n
    while(a[now]!=0):
        now=(now+1)%n
    k+=1;
    #print(k,end=" ")
    if(k==m):
        if(now==0):
            print(n,end=" ")
        else:
            print(now,end=" ")
        a[now]=1
        k=0
        c+=1
print()
ans=-1
for i in range(n):
    if(a[i]==0):
        ans=i
        break
if(ans==0):
    ans=n
print(ans)

C语言单向循环链表写法

#include <stdlib.h>
#include <stdio.h>

typedef struct _List {
	int num;
	struct _List *ne;
} List; //节点结构体

int n, m;
List *head, *tail; //头指针、尾指针

void init(int x) {
	List *temp = (List *)malloc(sizeof(List));
	temp -> num = x;
	temp -> ne = NULL;
	head = tail = temp;
} //初始化链表（第一个节点）

void insert_to_tail(int x) {
	List *temp = (List *)malloc(sizeof(List));
	temp -> num = x;
	temp -> ne = NULL;
	tail -> ne = temp;
	tail = temp;
} //将节点插入到尾部

int main()
{
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++) {
		if(i == 1) init(i);
		else insert_to_tail(i);
	} //建立链表
	tail -> ne = head; //首尾相连（循环链表）
	int rec = 0;
	for(List *temp = head, *pr = NULL; temp != temp -> ne;) { //遍历循环链表
		if(++rec % m == 0) { //踢人
			printf("%d ", temp -> num);
			pr -> ne = temp -> ne;
			free(temp);
			head = temp = pr -> ne;
		}
		else {
			pr = temp;
			head = temp = temp -> ne;	
		}
	}
	printf("\n%d\n", head -> num);
	return 0;	
}

using namespace std;
int n[10000];
int main(){
	int a,b;
	cin>>a>>b;
	for(int i=1;i<=a;i++){
		n[i]=i;
	}
	int geshu=0; 
	int k=0;
	for(int i=1;geshu!=a-1;){
		if(i==a+1){
			i=1;
		} 
		if(n[i]!=0)k++;
		if(k==b){
			cout<<n[i]<<" "; 
			n[i]=0;
			geshu++;
			k=0;
		}
		i++;
		
	}
	cout<<endl;
	for(int i=1;i<=a;i++){
		if(n[i]!=0){
			cout<<n[i];
			break;
		}
	}
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int a[10001];
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;++i)a[i]=false;
	int f=0,t=0,s=0;
	do{
		++t;
		if(t==n+1)t=1;
		if(a[t]==false)++s;
		if(s==m)
		{
			s=0;
			cout<<t<<" ";
			a[t]=true;
			++f;
		}
	}while(f!=n-1);
	for(int i=1;i<=n;++i)
	{
		if(a[i]==false)cout<<endl<<i;
	}
}

我特别喜欢某人说过的一句话：所有的约瑟夫问题都可以用队列或者线段树来解决，且思路简单于一般的数组模拟

#include <bits/stdc++.h>
using namespace std;

queue<int> q;
vector<int> ans;

int main()
{
	int n,m;
	cin >> n >> m;
	for(int i=1;i<=n;i++) q.push(i);
	int i=1;
	while(q.size()>1)
	{
		if(i == m)
		{
			int t = q.front();
			q.pop();
			ans.push_back(t);
			i = 1;
		}
		else
		{
			int t = q.front();
			q.pop();
			q.push(t);
			i++;
		}
	}
	for(auto x:ans) cout << x << " "; 
	cout << endl << q.front();
	return 0;
}

C++ :

#include <cstdio>
int main()
{
    printf("4");
    return 0;
}