tsinsen_a1014. Binary Conversion 5

The problem describes the given n hexadecimal positive integers, outputting their corresponding octal numbers. Enter the first behavior of the input format as a positive integer n (1<=n<=10).
Next n rows, each line a string of 0~9, uppercase letters A~F, representing the hexadecimal positive integer to be converted, each hexadecimal number is not more than 100000. The output format outputs n rows, and each behavior enters a corresponding octal positive integer. Note that the hexadecimal number entered does not have a leading 0, such as 012A.
The octal number of the output cannot have a leading 0. Sample Input 2
39
123ABC Sample Output 71
4435274 hints

The hexadecimal number is converted to a number of decimal digits, and then a binary number is converted into octal.

#include <stdio.h>//#include <iostream> #include <string.h>//using namespace std;//comment section was used to debug the const int MAXN = 400000 + 5;const int maxn2 = 100000 + 5;char s[maxn2];int D[MAXN], b[4], _count;int flag;void init () {Me Mset (S, ' n ', sizeof (s)), memset (d, 0, sizeof (d)), memset (b, 0, sizeof (b)); _count = 0;} int c_d (int ch) {if (ch >= ' a ' && ch <= ' F ') return (CH-' a ' +10); else return (ch-48);}   void Chu (int x, int* b, int cs) {int BCS = x, sz, cnt = 4;//memset (b, 0, sizeof (b));  Why this initialization does not work b[0] = b[1] = b[2] = b[3] = 0; This initialization succeeds while (SZ = (Bcs/cs)) {b[--cnt] = Bcs%cs;bcs = sz;//cout << "___" << cnt << "____" << Endl;}        B[--CNT] = Bcs%cs; The problem is it's out here. If the number of digits that are calculated does not meet 4 digits, you should deal with it.//int BZ = 3-cnt; BZ indicates that the insufficient number of digits is just the CNT value//cout << "insufficient digits" << BZ << Endl; /*for (int i = bz-1, j = 3; I >=0 && J >= BZ && cnt! = 0; I--, j--) {//shift number B[J-BZ] = b[cnt];cnt--; After the}//shift is over, fill all remaining slots with 0 for (inti = 3; I >= 4-bz; i--) B[i] = 0;*///This does not need to shift the operation,,, is their own thinking is not clear}int BQ (int x) {return 3-x%3;} int main () {int N, sum;while (scanf ("%d", &n)! = EOF) {for (int i = 0; i < N; i++) {init (); scanf ('%s ', s); for (int j = Strlen (s)-1; J >= 0; j--) {//memset (d, 0, sizeof (d)), flag = 0;chu (C_d (S[j]), B, 2),//cout << s[j] << "-----" << c_d (S[j]) &lt ;< "B[0"--> B[3] "<< b[0] <<" "<< b[1] <<" "<< b[2] <<" "<< b [3] << endl;for (int k = 3; k >= 0; k--)//Note I j K variables in these for loops {//cout << d[_count] << "* * * *"     ;d [_count++] = b[k]; cout << "<< b[i]" << b[i] << "_count" << _count << "" << D[_cou Nt-1] << ".  Above.. "<< Endl;}}  _count = _count + BQ (_count); Count the number of array elements required//cout << _count << "----___----" << endl;//for (int i = 0; i < _count; i++) cout << "<< d[i" << "";//cout &Lt;< Endl;int _cs = _count/3;for (int j = 0; J < _cs; J + +) {sum = d[_count-1]*4 + d[_count-2]*2 + d[_count-3];_count-   = 3;    printf ("%d", sum); Note that you cannot have a leading 0 or else you should use while to write a judgment to be determined! if (!flag &&!sum) continue;else {flag = 1; printf ("%d", sum);}} printf ("\ n");}} return 0;} My brother, I almost spit blood on this problem.

In the final analysis, I am too water ....

tsinsen_a1014. Binary Conversion 5