杭电acm—1376 Octal Fractions

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1376

Octal Fractions

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 832    Accepted Submission(s): 386

Problem Description

Fractions in octal (base 8) notation can be expressed exactly in decimal notation. For example, 0.75 in octal is 0.963125 (7/8 + 5/64) in decimal. All octal numbers of n digits to the right of the octal point can be expressed in no more than 3n decimal digits to the right of the decimal point.

Write a program to convert octal numerals between 0 and 1, inclusive, into equivalent decimal numerals. The input to your program will consist of octal numbers, one per line, to be converted. Each input number has the form 0.d1d2d3 ... dk, where the di are octal digits (0..7). There is no limit on k. Your output will consist of a sequence of lines of the form

0.d1d2d3 ... dk [8] = 0.D1D2D3 ... Dm [10]

where the left side is the input (in octal), and the right hand side the decimal (base 10) equivalent. There must be no trailing zeros, i.e. Dm is not equal to 0.

 

Sample Input

0.750.00010.01234567

 

Sample Output

0.75 [8] = 0.953125 [10]0.0001 [8] = 0.000244140625 [10]0.01234567 [8] = 0.020408093929290771484375 [10] 

 

Source

South Africa 2001

这一道题目，苦思冥想了很久，最后得到同学的点拨，一语惊醒梦中人，抓住一点，double是可以把后面的小数点计算出来的，只是输出的时候没有展现出来而已，所以，把小数点后的数字存储到字符数组里面就可以了。

AC代码如下：（如有错误和建议，请同学们不吝指出）

#include<stdio.h>#include<string.h>#include<math.h> char str[1000];char str10[1000];int main(){while(scanf("%s",str)!=EOF&&strcmp(str,"0")){int len=strlen(str);double sum=0;int k=1;for(int i=2;i<len;i++){sum+=(str[i]-'0')*1.0/pow(8,k++);}k=0;while(sum){//sum为0时退出 str10[k]=(int)(sum*10)%10+'0';//把小数点后的数字存到数组 sum=sum*10-(str10[k]-'0');//减去已存的数字 k++;}str10[k]='\0';//注意这个不可无 printf("%s [8] = 0.%s [10]\n",str,str10);}return 0;}