Hexadecimal to Octal
Thought analysis:
39 hexadecimal is first converted to 4-bit binary, 3 is 0011, 9 is 1001, so the binary of 39 is
00111001, and the binary is converted to 3-bit octal, counting from the right, 001 is 1, 111 is 7, so 39 The octal is 71.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_SIZE 100000
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
int main(int argc, char *argv[]) {
int n; //十六进制数数量
int i,j,k; // 计数变量
int result;
int str_len;
scanf("%d",&n);
char str[n][MAX_SIZE];
//建立数组，保存16个十六进制数对应的二进制数
char demo[16][4] = {
"0000","0001","0010","0011","0100","0101","0110","0111","1000","1001","1010","1011","1100","1101","1110","1111"};
memset(str,0,n*MAX_SIZE); //将str初始化为0
for(i=0;i<n;i++){
scanf("%s",str[i]);
k=0;
str_len = strlen(str[i]); //计数十六进制数长度
int ss = 4*str_len%3; //判断转换后的二进制数数量是否为的倍数，如果不是，需要补0
char temp[4*str_len+3-ss];
memset(temp,0,4*str_len+3-ss);
if(ss == 1){
strcat(temp,"00");
k += 2;
}
if(ss == 2){
strcat(temp,"0");
k +=1;
}
/* 进行十六进制向二进制数的转换 */
for(j=0;j<str_len;j++)
{
if(str[i][j]<65){
temp[k++] = demo[str[i][j]-48][0];
temp[k++] = demo[str[i][j]-48][1];
temp[k++] = demo[str[i][j]-48][2];
temp[k++] = demo[str[i][j]-48][3];
}
else if(str[i][j]>=65){
temp[k++] = demo[str[i][j]-55][0];
temp[k++] = demo[str[i][j]-55][1];
temp[k++] = demo[str[i][j]-55][2];
temp[k++] = demo[str[i][j]-55][3];
}
}
temp[k] = 0;
//进行输出操作
for(j=0;j<4*str_len;j+=3)
{
if(j==0){
result=(4*(temp[j]-48) + 2*(temp[j+1]-48) + (temp[j+2]-48));
if( result != 0)
printf("%d",result);
}
else{
result=(4*(temp[j]-48) + 2*(temp[j+1]-48) + (temp[j+2]-48));
printf("%d",result);
}
}
printf("\n");
}
return 0;
}