16 Conversion 8 Binary

It is mentioned in the title balloon that we can achieve the goal of 16 ext. 8 through the 2-step medium.

After learning to count the electrical logic, as we all know, 1-bit 16 binary can represent 4-bit 2-binary, 1-bit 8-binary can represent 3-bit binary, and for this reason my code big problem idea is out, string storage, 3-bit 16-sum into the stack output represents 4-bit 8-binary, and then out of the stack output, so I have the first code

 #include <iostream> #include <cstring> #include <cstdio> using namespace std; int
STACK[40000];
    void transform (string str, int length) {int top =-1;
        for (int i = length-1; I >= 0; I-= 3) {int sum = 0; for (int j = 0; J < 3 && i-j >= 0; j + +) {int temp = str[i-j] >= ' 0 ' && St R[I-J] <= ' 9 '?
            STR[I-J]-' 0 ': str[i-j]-' A ' + 10; sum + = (temp << (4 * j));
    Because each hexadecimal binary is accounted for in the binary 4 bits, so according to the position of the number is different, left 0 bits, 4 bits and 8 bits have reached his original decimal appearance} Stack[++top] = sum;
    } while (stack[top] = = 0) {top--; 
    } for (int i = top; I >= 0; i--) {printf ("%o", Stack[i]);
} cout<<endl;
	} int Main () {string *str;
	int n;
	cin>>n;
	str = new String[n];
	for (int i = 0; i < n; i++) {cin>>str[i];
	} for (int i = 0; i < n; i++) {transform (Str[i], str[i].size ());
} return 0; }

But unfortunately, in the evaluation platform is 0 points, I have tried some data with others, but the test is basically not a string of f is a string of other, but it is to help people find out a mistake, but their mistakes still did not find, once thought is my stack open is not big enough, so the last data overflow, cannot be expressed accurately. Later consulted the C Language bar goddess of the smokers flame I found my Code error place.  The wrong place lies in the lazy printf ("%o", Stack[i]); Because I am lazy sums, for this lost many of the leading 0.

The correct result after 1001 conversion should be 10001.

In order to correct this error, I finally converted to a string to see if the current number is enough to 4 bits, for this there is the following code.

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int stack[40000];
void transform (string str, int length)
{
    char buff[4]; 
    int top =-1;
    for (int i = length-1; I >= 0; I-= 3)
    {
        int sum = 0;
        for (int j = 0; J < 3 && i-j >= 0; j + +)
        {
            int temp = str[i-j] >= ' 0 ' && Str[i-j] & lt;= ' 9 '? STR[I-J]-' 0 ': str[i-j]-' A ' + ten;
            sum + = (temp << (4 * j));
        }
        Stack[++top] = sum;
    }
    while (stack[top] = = 0)
    {
    	top--;
    }
    for (int i = top; I >= 0; i--)
    {
        printf ("%04o", Stack[i]);
    }
    cout<<endl;
}

int main ()
{
	string *str;
	int n;
	cin>>n;
	str = new String[n];
	for (int i = 0; i < n; i++)
	{
		cin>>str[i];
	}
	for (int i = 0; i < n; i++)
	{
		transform (str[i], str[i].size ());
	}
	
	return 0;
}

This will allow the answer to be output correctly.