进制转换 - 数据结构与算法

已结贴√ 问题点数：20 回复次数：3

进制转换

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define STACK_INIT_SIZE 20
#define STACKINCREMENT  10

typedef char ElemType;
typedef struct
{
ElemType *base;
ElemType *top;
int stackSize;
}sqStack;

void InitStack(sqStack *s)
{
s->base = (ElemType *)malloc(STACK_INIT_SIZE * sizeof(ElemType));
if( !s->base )
    {
exit(0);
    }

s->top = s->base;
s->stackSize = STACK_INIT_SIZE;
}

void Push(sqStack *s, ElemType e)
{
if( s->top - s->base >= s->stackSize )
    {
s->base = (ElemType *)realloc(s->base, (s->stackSize + STACKINCREMENT) * sizeof(ElemType));
if( !s->base )
        {
exit(0);
        }
    }

    *(s->top) = e;
s->top++;
}

void Pop(sqStack *s, ElemType *e)
{
if( s->top == s->base )
    {
return;
    }
    *e = *--(s->top);
}

int StackLen(sqStack s)
{
return (s.top - s.base);
}

int main()
{
ElemType c;
sqStack s;
int len, i, sum = 0;

InitStack(&s);

printf("请输入二进制数，输入#符号表示结束！\n");
scanf("%c", &c);
while( c != '#' )
    {
Push(&s, c);
scanf("%c", &c);
    }

getchar();  // 把'\n'从缓冲区去掉

len = StackLen(s);
printf("栈的当前容量是: %d\n", len);

for(i=0; i<len; i++ )
    {
Pop(&s, &c);
sum = sum + (c-48) * pow(2, i);
    }

printf("转化为十进制数是: %d\n", sum);

return 0;
}
这是一个二进制转换为十进制的代码，有没有什么更好的办法改为二进制转换为八进制？

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第 2 楼

得分:14

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define STACK_INIT_SIZE 20
#define STACKINCREMENT  10

typedef char ElemType;
typedef struct
{
ElemType *base;
ElemType *top;
int stackSize;
}sqStack;

void InitStack(sqStack *s)
{
s->base = (ElemType *)malloc(STACK_INIT_SIZE * sizeof(ElemType));
if( !s->base )
    {
exit(0);
    }

s->top = s->base;
s->stackSize = STACK_INIT_SIZE;
}

void Push(sqStack *s, ElemType e)
{
if( s->top - s->base >= s->stackSize )
    {
s->base = (ElemType *)realloc(s->base, (s->stackSize + STACKINCREMENT) * sizeof(ElemType));
if( !s->base )
        {
exit(0);
        }
    }

    *(s->top) = e;
s->top++;
}

void Pop(sqStack *s, ElemType *e)
{
if( s->top == s->base )
    {
return;
    }
    *e = *--(s->top);
}

int StackLen(sqStack s)
{
return (s.top - s.base);
}
void f_8(int n){
    int m=n%8;n/=8;
    if(n)f_8(n);printf("%d",m);
}
int main()
{
ElemType c;
sqStack s;
int len, i, sum = 0;

InitStack(&s);

printf("请输入二进制数，输入#符号表示结束！\n");
scanf("%c", &c);
while( c != '#' )
    {
Push(&s, c);
scanf("%c", &c);
    }

getchar();  // 把'\n'从缓冲区去掉

len = StackLen(s);
printf("栈的当前容量是: %d\n", len);

for(i=0; i<len; i++ )
    {
Pop(&s, &c);
sum = sum + (c-48) * pow(2, i);
    }

printf("转化为十进制数是: %d\n", sum);
printf("转化为八进制数是: ");
f_8(sum);
return 0;
}

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第 3 楼

得分:0

回复 2楼林月儿

我也是这样想的，通过一个函数把输出的十进制转化为八进制，但是它的要求是利用栈，直接把二进制转化为八进制

第 4 楼

得分:0

那就main函数改成,楼主排版。。。
int main()
{
ElemType c;
sqStack s;
int len, i, sum = 0;

InitStack(&s);

printf("请输入二进制数，输入#符号表示结束！\n");
scanf("%c", &c);
while( c != '#' )
    {
Push(&s, c);
scanf("%c", &c);
    }

getchar();  // 把'\n'从缓冲区去掉

len = StackLen(s);
printf("栈的当前容量是: %d\n", len);

for(i=0; i<len; i++ ){
Pop(&s, &c);
sum = sum + (c-48) * pow(2, i%3);
if(i%3==0){
    printf("%d",sum);
    sum=0;
}
}
return 0;
}

剑栈风樯各苦辛，别时冰雪到时春