蓝桥杯 基础练习 十六进制转八进制

#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD doublea
#define MAX 100100
#define mod 10007
using namespace std;
char a[MAX];
int p[MAX*10];
int main()
{
    int n,m,j,i,len,sum;
    scanf("%d",&n);
    while(n--)
    {   	
    	scanf("%s",&a);	
    	char s[MAX*10]="";
    	len=strlen(a);
    	for(i=0;i<len;i++)  //转化为二进制 
    	{
    	    switch(a[i])
    	    {
    	    	case ‘0‘:strcat(s,"0000");break;
    	    	case ‘1‘:strcat(s,"0001");break;
    	    	case ‘2‘:strcat(s,"0010");break;
    	    	case ‘3‘:strcat(s,"0011");break;
    	    	case ‘4‘:strcat(s,"0100");break;
    	    	case ‘5‘:strcat(s,"0101");break;
    	    	case ‘6‘:strcat(s,"0110");break;
    	    	case ‘7‘:strcat(s,"0111");break;
    	    	case ‘8‘:strcat(s,"1000");break;
    	    	case ‘9‘:strcat(s,"1001");break;
    	    	case ‘A‘:strcat(s,"1010");break;
    	    	case ‘B‘:strcat(s,"1011");break;
    	    	case ‘C‘:strcat(s,"1100");break;
    	    	case ‘D‘:strcat(s,"1101");break;
    	    	case ‘E‘:strcat(s,"1110");break;
    	    	case ‘F‘:strcat(s,"1111");break;
    	    	default:break;
    	    }
    	}
    	//puts(s);
    	strrev(s);
    	int len1=strlen(s);
    	if(len1%3==1) //因为二进制转八进制，是有三位二进制变为一位八进制 
    	{           //所以要补全，在前边补0使长度能整除3 
    		strcat(s,"00");
    		len1+=2;
    	}
    	else if(len1%3==2)
    	{
    		strcat(s,"0");
    		len1+=1;
    	}
    	int op=1;
    	for(i=0;i<len1-2;i+=3)  //二进制转八进制 
    	{
    		p[op++]=(s[i]-‘0‘)*1+(s[i+1]-‘0‘)*2+(s[i+2]-‘0‘)*4;
    	}
    	int flag=1;
    	for(i=op;i>0;i--)
    	{
    		if(p[i]!=0)
    		    flag=0;
    		if(p[i]==0&&flag)
    		    continue;
    		printf("%d",p[i]);
    	} 
        printf("\n");
    }
	return 0;
} 

#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD doublea
#define MAX 100100
#define mod 10007
using namespace std;
int p[MAX*10];
int main()
{
    int n,m,j,i,len,sum;
    scanf("%d",&n);
    while(n--)
    {
    	string a,s;
    	a.resize(MAX); //需要预先分配空间 
    	scanf("%s",&a[0]);
    	s="";
    	len=a.length();//字符串长度 
    	for(i=0;i<len;i++)
    	{
    	    switch(a[i])
    	    {
    	    	case ‘0‘:s+="0000";break;
    	    	case ‘1‘:s+="0001";break;
    	    	case ‘2‘:s+="0010";break;
    	    	case ‘3‘:s+="0011";break;
    	    	case ‘4‘:s+="0100";break;
    	    	case ‘5‘:s+="0101";break;
    	    	case ‘6‘:s+="0110";break;
    	    	case ‘7‘:s+="0111";break;
    	    	case ‘8‘:s+="1000";break;
    	    	case ‘9‘:s+="1001";break;
    	    	case ‘A‘:s+="1010";break;
    	    	case ‘B‘:s+="1011";break;
    	    	case ‘C‘:s+="1100";break;
    	    	case ‘D‘:s+="1101";break;
    	    	case ‘E‘:s+="1110";break;
    	    	case ‘F‘:s+="1111";break;
    	    	default:break;
    	    }
    	}
    	//puts(s);
    	//strrev(s);
    	int len1=s.length();
    	if(len1%3==1)
    	{
    		s="00"+s;  //00+s是将00加在字符串前  s+00是将00加在字符串后 
    		len1+=2;
    	}
    	else if(len1%3==2)
    	{
    		s="0"+s;
    		len1+=1;
    	}
    	int op=1;
    	for(i=0;i<len1-2;i+=3)
    	{
    		p[op++]=(s[i]-‘0‘)*4+(s[i+1]-‘0‘)*2+(s[i+2]-‘0‘)*1;
    	}
    	int flag=1;
    	for(i=0;i<op;i++)
    	{
    		if(p[i]!=0)
    		    flag=0;
    		if(p[i]==0&&flag)
    		    continue;
    		printf("%d",p[i]);
    	} 
        printf("\n");
    }
	return 0;
}