C language solves the Josephian problem, how should C solve the Josephian problem ?

I wrote it myself and solved it directly with a one-dimensional array.

#include

define n 9 total number of people.

define k 1 to start counting.

define m 3 intervals.

Assign a value to the array.

void setdate(int a,int n)int i;

for(i=0;i=len) dm=dm-len;

printf("The last child was %d",a[0]);The last child is placed in a[0].

main()

int a[n];

setdate(a,n);

play(a,k,m);

#include

#include

typedef struct node

int key;

int data;

struct node *next;

node,*linklist;

void initlist(linklist *l);

void creat(linklist l,int n);

void find(linklist l,int n);

void main()

linklist l;

int n;

printf("Please enter the number of people:");

scanf("%d",&n);

initlist(&l);

printf("Enter each person's password in turn");

creat(l,n);

printf("Please enter your initial password");

find(l,n);

void initlist(linklist *l)l=(linklist)malloc(sizeof(node));

l)->next=null;

void creat(linklist l,int n)node *s,*r;

int i=1,t;

r=l;while(i<=n)

scanf("%d",&t);

s=(node*)malloc(sizeof(node));

s->key=t;

s->data=i;

r->next=s;

r=s;i++;

r->next=l->next;

void find(linklist l,int n)int n,i=1,t,j=1;

node *p,*q;

p=l;scanf("%d",&n);

t=n;printf("The order of the columns is:");

while(j<=n)

while(inext;

printf("%d ",p->next->data);

q=p->next;

t=p->next->key;

p->next=q->next;

i=1;j++;

printf("");

This is the Joseph loop of the data structure, you see, just modify it...

1. To solve this problem, you can use the method of simulating the number to create an array of size n, the nth element of the array indicates whether the nth person is still in the team, first set each element to 1, indicating that all members are in the team. If the nth person is on the team, the nth element is set to 0.

2. You can use an accumulation counter to simulate the number of reports, which indicates how many people have reported the number of reports in this round, and then iterate through each person, if they are in the team, the counter will be +1, if the number is accumulated to m, then this person is out of the team. And so on until n-1 person comes out of the team, and only 1 person remains.

3. Finally, iterate through the array and find the people who are still in the team.