1004: Octal Fractions 利用秦九韶定理进行小数的进制转换

本文链接：https://blog.csdn.net/kongming_acm/article/details/5802694

该程序旨在将0到1范围内的八进制小数转换为等价的十进制小数。输入为形式为0.d1d2...dk的八进制数，输出为无尾随零的十进制表示。

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1004: Octal Fractions

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	3535	790	Standard

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

0.75
0.0001
0.01234567

SAMPLE OUTPUT

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

分析：0.12345【8】=1*1/8+2*1/(8*8）+3*1/(8*8*8)+4*1/(8*8*8*8)+5*1/(8*8*8*8*8)
=1000^5[1*1/8+2*1/(8*8）+3*1/(8*8*8)+4*1/(8*8*8*8)+5*1/(8*8*8*8*8)]/1000^5
=(1*125*1000^4+2*125^2*1000^3+3*125^3*1000^2+4*125^4*1000+5*125^5)/1000^5

=(1*125*1000^4+2*125^2*1000^3+3*125^3*1000^2+125^4*(4*1000+（5*125)）)/1000^5
=(1*125*1000^4+2*125^2*1000^3+125^3*(3*1000^2+125*(4*1000+（5*125)）))/1000^5
=(1*125*1000^4+125^2*(2*1000^3+125*(3*1000^2+125*(4*1000+（5*125）))))/1000^5
=125*(1*1000^4+125*(2*1000^3+125*(3*1000^2+125*(4*1000+（5*125）)))))/1000^5

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;
const int maxn=1000;
int a[maxn+1];
int main()
{
    string str;
    while(cin>>str)
    {
        int n=str.size()-2;
        memset(a,0,sizeof(a));
        int t=maxn;
        for(int i=str.size()-1;i>=2;i--,t-=3)
        {
            a[t]+=str[i]-'0';
            for(int i=t;i>=0;i--)
            {
                if(a[i]>=10) a[i]%=10,a[i-1]++;
                else break;
            }
            for(int i=maxn;i>=0;i--) a[i]*=125;
            for(int i=maxn;i>=0;i--)
            {
                if(a[i]>=10) a[i-1]+=a[i]/10,a[i]%=10;
            }
        }
        int m;
        for(m=maxn;a[m]==0;m--);//去掉尾部的0
        cout<<str<<" [8] = 0.";
        for(int i=maxn-3*n+1;i<=m;i++) cout<<a[i];
        cout<<" [10]"<<endl;
    }
    return 0;
}