Using structures and arrays to implement Joseph Ring (second of the problem of ring counting)

The day before yesterday with a single cycle linked list to achieve the Joseph ring problem, this method of implementation efficiency. This is followed by two other simple methods.

Method one: Using arrays

　　

voidMain () {inta[Bayi],n,i,counter,num; Counter is used to calculate, num is used to record the number of exits printf ("Please input all the number of person :"); scanf ("%d",&N);  for(i=0; i<n;i++)//number for each person, 1~n {A[i]=i+1; }     for(i=0; i<n;i++)//print pre-status printf ("%d\t", A[i]); printf ("\ n"); Counter=0; i=0; num=0;  while(num< (n1) &&i<N) {if(a[i]!=0)//When the number is not 0 o'clock, the value plus 1 counter++; if((counter%3)==0&&a[i]!=0)//When the count value can be divisible by 3 and the element number is not 0 o'clock, the number value is placed 0,num plus 1****** this place at first did not judge A[i] is not equal to 0,
The result is a logic error {printf ("Number %d is out\n", A[i]); A[i]=0; Num++; if(num==n-1) Break; If NUM is the total number minus 1, then there is still one person. Roll out loop} I++; Loop Arrayif(i==N)//When the count is up to the last one, start the count again {i=0; }} printf ("\nthe last person is number"); Number of people remaining after the print operation for(i=0; i<n;i++)        if(a[i]!=0) printf ("%d\t", A[i]);}

Method two: Using struct arrays

　　

structperson{//Each element defines a member that holds the subscript nextp of the next element, a member holds the member numberintNumber ; intnextp;};voidMain () {intN out, Num,i; printf ("Please input the total number of the people:"); scanf ("%d",&N); structPerson Link[n];  for(i=0; i<n;i++) {if(i==n-1) link[i].nextp=0; The last person's next element position is the first element, forming a loop queueElselink[i].nextp=i+1; The next person's serial number Link[i].number=i+1; My number} printf ("\ n");  out=0; num=n-1; Define the number of out-logged exits, initialize num as the last element to make it easier to count off the first element while( out<n-1) {i=0;  while(i!=3) {num=LINK[NUM].NEXTP; Count off from the first elementif(Link[num].number) i++; When the number is not 0 o'clock, the Count plus 1} printf ("Number %d is out\n", Link[num].number); Print out the exit personnel number Link[num].number=0; Exit Person number 0 out++; } printf ("\nthe last person is number");  for(i=0; i<n;i++)        if(link[i].number!=0) printf ("%d\t", Link[i].number);}

　

Using structures and arrays to implement Joseph Ring (second of the problem of ring counting)