Use the stack to convert the base system, how many base systems does the stack have?

Recursive (using stacks automatically assigned by the system):

#include

using namespace std;

#define maxn 1000

bool isok(char *n)

int len = strlen(n);

for(int i=0; i < len; i++)if(n[i] != '0') return false;

return true;

void dec2bin(char* n)if(!isok(n))

int ys = 0;

int len = strlen(n);

int ans = (n[len - 1]-'0') %2;

for(int i=0; i

n;int len = strlen(n);

dec2bin(n);

cout non-recursive** (although it is not written in a normal way, ans is a stack):

#include

using namespace std;

#define maxn 10000

bool isend(char *n)

for(int i=0; i

0) return false;

return true;

int dec2bin()

char n[maxn];

char ans[maxn];

memset(n, 0, sizeof(n));

memset(ans, 0, sizeof(ans));

cin >>n;

int len = strlen(n);

int ans_i = 0;

for(int i=0; i

0; i--)

cout int main()

while(1)

dec2bin();

return 0;

This has to do with the algorithm of the base conversion. For example:

56d to 3 is converted to base system, and the specific algorithm is as follows.

The integer working variable a is stored for 56d

When a is not equal to 0, start the loop: calculate a%3, get the remainder 2, this is the lowest bit of the result (the 0th digit), save, calculate a=a 3=18

When a is not equal to 0, the cycle continues: calculate a%3, get the remainder 0, this is the first digit of the result, save, calculate a=a 3=6

When a is not equal to 0, the cycle continues: calculate a%3 to get the remainder 0, this is the second digit of the result, save, calculate a=a 3=2

When a is not equal to 0, the cycle continues: calculate a%3, get the remainder 2, this is the 3rd digit of the result, save, calculate a=a 3=0

If the cycle condition of a is not equal to 0 is not satisfied, the cycle is stopped.

At this point, we can see that the remainder we save is in reverse order, and to compose the final number, we need to output it in reverse order.

The stack happens to have a first-in, first-out feature, so we press the above remainders into the stack in order, and when the loop is over, all the remainders out of the stack will get the output in the correct order. (You can save it as a stack in the above steps).

Data structures, stack applications, can be converted to base through subroutines or functions, can be programmed into a word program, or a custom function can be used to convert binary, decimal, hexadecimal, etc.

There are a total of 24 kinds of full arrangement of the 4 elements, and the stack requirements meet the last-in-first-out

14 possibilities, 10 impossible.