queue or 数组模拟

#include <bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef pair<int, int> PII;
using ll = long long;
using ULL = unsigned long long;
const int N = 1e7 + 5;

int n, k, a[N];
//queue<int> q;
int q[N], hh = 0, tt = -1;
inline void solve() {
/*
    cin >> n >> k;
    for (int i = 1; i <= n; i++) q.push(i);
    ll cnt = 0, id = 0;
    while (!q.empty()) {
        int hh = q.front();
        q.pop();
        cnt++;
        if (cnt == k) {
            id++;
            a[hh] = id;
            cnt = 0;
        }else q.push(hh);
    } 
    for (int i = 1; i <= n; i++) cout << a[i] << " \n"[i==n];
*/
    cin >> n >> k;
    for (int i = 1; i <= n; i++) q[++tt] = i;
    int cnt = 0, idx = 0;
    while (hh <= tt) {
        int aa = q[hh];
        ++hh;
        cnt++;
        if (cnt == k) {
            idx++;
            a[aa] = idx;
            cnt = 0;
        }else q[++tt] = aa;
    }
   for (int i = 1; i <= n; i++) cout << a[i] << " \n"[i==n];
}

int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int _ = 1;
    //int _; cin >> _;
    while (_--) solve();
    return 0;
}

#include<iostream>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;
using ll = long long;
const int N = 1e3+10;
ll a[N];
int main()
{
    int n,k;cin>>n>>k;
    int idx = 0;
    int cnt = 0;
    queue<int> q;

    for(int i = 1;i <= n;i++)
    {
        q.push(i);
    }

    while(!q.empty())
    {
        int hh = q.front();
        q.pop();
        cnt ++;
        if(cnt == k)
        {
            idx++;
            a[hh] = idx;
            cnt = 0;
        }else q.push(hh);
    }
    for(int i = 1;i <= n;i++)cout<<a[i]<<" ";
   
    return 0; 
}

c

#include<stdio.h> int main() { int m,n; int temp[1000]; int t=0;//记第几个死 int f[1000]; scanf("%d %d",&m,&n); for(int i=0;i<m;i++) { f[i]=i+1; }//标号每个人的站位 int i=0; int k=0;//报数 int c=m;//表示还剩的人数 while(c>0)//死光了就不循环了 { if(f[i]!=0)//死人不用报数 k++;//开始报数 if(k==n) { t++; temp[i]=t; c--; f[i]=0;//死人腾出站位 k=0;//重新报数 } i=(i+1)%m;//围成了圈，当i=m-1时，i=0，又重新开始数
} for(int j=0;j<m;j++) printf("%d ",temp[j]); return 0; }

这道题是一道经典的循环链表

#include <bits/stdc++.h>

//循环链表

using namespace std;

const int maxn = 1e3 + 5;
int id[maxn];

typedef struct Node {
int value;
Node *next;
}NodeList;

int main(void) {
int n, m;
cin >> n >> m;
NodeList *head, *last = NULL;
NodeList *node;

for (int i = 1; i <= n; i++) {
    node = (NodeList*)malloc(sizeof(NodeList));
    node->value = i;
    if (i == 1) {
        head = node;
        last = node;
    }else {
        last->next = node;
        last = node;
    }
}
node->next = head;
last = node;

int i = 1;
int cnt = 1;
while (head->next != head) {
    if (i == m) {
        NodeList *p = head;
        last->next = head->next;
        id[p->value] = cnt;
        cnt ++;
        free(p);
        i = 1;
        head = last->next;
    }else {
        last = head;
        head = head->next;
        i ++;
    }
}

for (int i = 1; i <= n; i++) {
    if (!id[i]) id[i] = n;
    cout << id[i] << ' ';
}

return 0;

}

约瑟夫环 经典题了，用 队列 随别 虐杀。数组的话 也可以。

#include <iostream>
#include <queue>

using namespace std;

int arr[1005]; // 到底是 第几次 自杀的 都存在 这里

int main(void) {
	int n, k;// n 个 人，报数到 k 就会自杀
	cin >> n >> k;

	queue<int> que;

	for (int i = 1; i <= n; ++i) {
		que.push(i); // 先把这些人 都添加到 队列里
	}

	int ans = 0;// 记录是第几个 出去的
	int jilu = 0; // 记录 数到 几了
	while (que.size()) {
		int id = que.front();// 编号
		que.pop();
		jilu++;
		if (jilu != k) {
			que.push(id);
		}
		else {
			ans++;
			arr[id] = ans;
			jilu = 0;
		}
	}

	for (int i = 1; i <= n; ++i) {
		cout << arr[i] << " ";
	}

	return 0;
}

#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <vector>
#include <iostream>
#include <string.h>
#include <queue>
using namespace std;
int n,k;
queue<int> q;
int a[1005];
int main() {
    scanf("%d%d",&n,&k);
    for(int i=1; i<=n; i++) {
        q.push(i);
    }
    int sum=1,i=0,j=0;
    while(sum<n) {
        if(j!=k-1) {
            int tmp=q.front();
            q.pop();
            q.push(tmp);
            j++;
        } else {
            a[q.front()]=sum;
            sum++;
            q.pop();
            j=0;
        }
    }
    a[q.front()]=sum;
    for(int i=1; i<=n; i++)
        printf("%d ",a[i]);
    return 0;
}