POJ – 1131

Octal Fractions

Submit Status

Description

Fractions in octal (base 8) notation can be expressed exactly in decimal notation. For example, 0.75 in octal is 0.953125 (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.

Input

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.

Output

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

0.75
0.0001
0.01234567

Sample Output

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

特别的神奇。。。。。。

啊啊。。。

从末位往前算，每次相当于把已经算出的数字往后退了一位

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>

using namespace std;

#define For(i, a, b) for(int i = a; i <= b; i++)
#define Forr(i, a, b) for(int i = a; i >= b; i--)
#define MAXN (100+5)

char s8[MAXN];

int main(){
//	freopen("test.in", "r", stdin);
//	freopen("test.out", "w", stdout);

	while(scanf("%s", s8) != EOF){
		char s10[MAXN] = {'0'};
		int len8 = strlen(s8), end = 0;
		
		Forr(i, len8-1, 2){
			int j, num = s8[i]-'0';
			for(j = 0; (j < end) || num; j++){
				int tmp = num*10 + (j<end? s10[j]-'0': 0);
				s10[j] = tmp/8 + '0';
				num = tmp % 8;
			}
			end = j;
		}
		s10[end] = '\0';

		printf("%s [%d] = 0.%s [%d]\n", s8, 8, s10, 10);
	}

	return 0;
}