第1个回答 2012-12-26
#include <stdio.h>
//获取字符串长度
int getLength(char a[])
{
int count=0;
while (a[count]) {
count++;
}
return count;
}
//反转字符串
void reverse(char a[])
{
char temp;
int length=getLength(a), i=0;
while (i<length/2) {
temp=a[i];
a[i]=a[length-i-1];
a[length-i-1]=temp;
i++;
}
}
//base的count次方
int intPow(int base,int count)
{
int index=0,result=1;
while (index<count) {
result=result*base;
index++;
}
return result;
}
//以base进制将数字转化为字符串,
//如10对应十进制{'1','0'}对应8进制{'1','2'}
void convertIntToChars(int a,int base,char result[])
{
int quotient=a,index=0;
while (quotient) {
result[index]=quotient%base+'0';
quotient=quotient/base;
index++;
}
result[index]=0;
reverse(result);
}
//将字符串以base进制转换为10进制
int atoi(char a[],int base)
{
int length=getLength(a),sum=0,index=0;
while(index<length)
{
sum+=(a[index]-'0')*intPow(base,length-1-index);
index++;
}
return sum;
}
//十进制转八进制,并以十进制形式表示,如8(十进制)对应10(八进制)
int decimalToOctal(int a)
{
char result[100];
convertIntToChars(a,8,result);
returnatoi(result,10);
}
//八进制转十进制,并以十进制形式表示
int octalToDecimal(int a)
{
char result[100];
convertIntToChars(a,10,result);
returnatoi(result,8);
}
int main(int argc, constchar * argv[])
{
int decimal=10;
printf("decimal %d's octal number is %d\n",decimal,decimalToOctal(decimal));
int octal=10;
printf("ocatal %d's decimal number is %d\n",octal,octalToDecimal(octal));
return0;
}
第2个回答 2012-12-26
试一下以下程序段:
int a[32]={};
int x,i=-1;
scanf("%d",&x);
while (x>0)
{
a[++i]=x % 10;
x/=10;
}
int s=0;
while (i>=0)
{
s=s*10+a[i--];
}
printf("(%d)\n",s);