Void main () 
{
Int people [1000], n_5; 
Printf (" both Please enter the number of people: \ n "); 
The scanf (" % d ", & amp; N_5); 

for(int i=0; i{
People [I]=I + 1; 
} 
//printf (" people [4]=% d \ n ", people [4]); 
Int * p_5=people; 
Int remain=n_5 num_off=1; 
While (remain> 1)//the rest of the Numbers is greater than 1 
{

If ((p_5) & lt;=(people + n_5-1))//the man is not a number of the last 
{
//printf (" p_5=% d ", p_5); 

While ((* p_5)!=0)//this person not registered before 3 
{

If (num_off!=3)//who count off for 3 
{

Num_off + +; 

} 
Else if (num_off==3)//who count off for 3 
{

* p_5=0; 
Remain -; 

Num_off=1; 
} 

Printf (" * p_5 num_off==% d % d \ n ", * p_5, num_off);//for observation process 
P_5 + +; 

} 

} 



Else if (p_5 & gt; (people + n_5-1)) 
{
P_5=people; 
While ((* p_5)!=0) 
{
If (num_off! 
=3){
Num_off + +; 

} 
Else if (num_off==3) 
{
* p_5=0; 
Remain -; 

Num_off=1; 
} 
P_5=people; 
} 


} 
} 

While ((* p_5)==0) 
{

P_5 + +; 
} 
Printf (" The number of remained one is: % d \ n ", * p_5); 
} 

Here is the run results, trouble to help have a look, thank!

CodePudding user response:

Suggest their abilities of debugging and coding debugged, equally important, even more important

CodePudding user response:

Fyi:

//assume there are n people completely surrounded, personal start counting from the first, at the first m personal time, the first m personal dequeue, 
//and then continue to count from 1 began to count, to the first m personal exit 
#include  
#include  
Int I, k, t; 
int n,m; 
The static char f [1001];//0 the seats are not ones, 1 the seat has been ones with 
Void main () {
While (1) {
Printf (" Input n m (1000 & gt;=n>=m>=1) : "); 
The fflush (stdout); 
The rewind (stdin); 
If (2==the scanf (" % d % d ", & amp; N, & amp; M)) {
If (1000 & gt;=n & amp; & N>=m & amp; & M>=1) break; 
} 
} 
T=0;//have ones with the total number of 
i=1;//seat number 
k=1;//the current wants to count the number of 
While (1) {
If (==0 f [I]) {
If (m==k) {
T++; 
F [I]=1; 
Printf (" % 3 d ", I); 
If (0==t % 10) printf (" \ n "); 
If (t>=n) break; 
} 
k++; If (k> M), k=1; 
} 
i++; If (i> N) I=1; 
} 
Cprintf (" Press any key... "); 
getch(); 
} 

CodePudding user response:

Reference:

 void main () 
{
Int people [1000], n_5; 
Printf (" both Please enter the number of people: \ n "); 
The scanf (" % d ", & amp; N_5); 

for(int i=0; i{
People [I]=I + 1; 
} 

Int * p_5=people; 
Int remain=n_5 num_off=0; 
While (remain> 1)//the rest of the Numbers is greater than 1 
{
P_5=people;//here is equivalent to the end to end of an array is 
While ((p_5) & lt;=(people + n_5-1))//determine whether in the array, to tail jump out to a 
{//this cycle is the length of a n 

If (* p_5!=0)//judgment to have ones, if you have ones with the skip, enter p_5 + + a data judgment 
{
Num_off + +;//not ones to count 1 

If (num_off==3) 
{
Printf (" this is a registration number 3: * p_5 num_off==% d % d \ n ", * p_5, num_off);//for observation process 
* p_5=0; 
Remain -;//the total number of minus one 
Num_off=0;//start from 0 to count 
} 
} 
//printf (" * p_5 num_off==% d % d \ n ", * p_5, num_off);//for observation process 
P_5 + +;//array under a data 
} 
} 

P_5=people;//right now to make p_5 array pointer back to the first position 

While (* p_5==0)//find the location of the last is not zero 
P_5 + +; 
Printf (" The number of remained one is: % d \ n ", * p_5); 

} 

CodePudding user response:

reference 1st floor flying_music response:

suggest their debugging, debugging with coding ability is just as important, even more important

Know, thank you! This is already a lot of times, or not??

CodePudding user response:

refer to the second floor 4 teacher zhao response:

are for reference only:

//assume there are n people completely surrounded, personal start counting from the first, at the first m personal time, the first m personal dequeue, 
//and then continue to count from 1 began to count, to the first m personal exit 
#include  
#include  
Int I, k, t; 
int n,m; 
The static char f [1001];//0 the seats are not ones, 1 the seat has been ones with 
Void main () {
While (1) {
Printf (" Input n m (1000 & gt;=n>=m>=1) : "); 
The fflush (stdout); 
The rewind (stdin); 
If (2==the scanf (" % d % d ", & amp; N, & amp; M)) {
If (1000 & gt;=n & amp; & N>=m & amp; & M>=1) break; 
} 
} 
T=0;//have ones with the total number of 
i=1;//seat number 
k=1;//the current wants to count the number of 
While (1) {
If (==0 f [I]) {
If (m==k) {
T++; 
F [I]=1; 
Printf (" % 3 d ", I); 
If (0==t % 10) printf (" \ n "); 
If (t>=n) break; 
} 
k++; If (k> M), k=1; 
} 
i++; If (i> N) I=1; 
} 
Cprintf (" Press any key... "); 
getch(); 
} 

Ok, thank you!

CodePudding user response:

The

reference 3 floor QZJHJXJ response:

for reference:

 void main () 
{
Int people [1000], n_5; 
Printf (" both Please enter the number of people: \ n "); 
The scanf (" % d ", & amp; N_5); 

for(int i=0; i{
People [I]=I + 1; 
} 

Int * p_5=people; 
Int remain=n_5 num_off=0; 
While (remain> 1)//the rest of the Numbers is greater than 1 
{
P_5=people;//here is equivalent to the end to end of an array is 
While ((p_5) & lt;=(people + n_5-1))//determine whether in the array, to tail jump out to a 
nullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnullnull