POJ - 1930 Dead Fraction（数学精度计算）

最新推荐文章于 2021-09-25 23:32:34 发布

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最新推荐文章于 2021-09-25 23:32:34 发布

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本文针对POJ 1930 DeadFraction问题进行解析，介绍了一种快速找出最简分数的方法，该方法能从给定的近似小数中恢复出最简单的分数形式。

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题目链接：http://poj.org/problem?id=1930点击打开链接

Dead Fraction

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 3242		Accepted: 1068

Description

Mike is frantically scrambling to finish his thesis at the last minute. He needs to assemble all his research notes into vaguely coherent form in the next 3 days. Unfortunately, he notices that he had been extremely sloppy in his calculations. Whenever he needed to perform arithmetic, he just plugged it into a calculator and scribbled down as much of the answer as he felt was relevant. Whenever a repeating fraction was displayed, Mike simply reccorded the first few digits followed by "...". For instance, instead of "1/3" he might have written down "0.3333...". Unfortunately, his results require exact fractions! He doesn't have time to redo every calculation, so he needs you to write a program (and FAST!) to automatically deduce the original fractions.
To make this tenable, he assumes that the original fraction is always the simplest one that produces the given sequence of digits; by simplest, he means the the one with smallest denominator. Also, he assumes that he did not neglect to write down important digits; no digit from the repeating portion of the decimal expansion was left unrecorded (even if this repeating portion was all zeroes).

Input

There are several test cases. For each test case there is one line of input of the form "0.dddd..." where dddd is a string of 1 to 9 digits, not all zero. A line containing 0 follows the last case.

Output

For each case, output the original fraction.

Sample Input

0.2...
0.20...
0.474612399...
0

Sample Output

2/9
1/5
1186531/2500000

后面的小数点不一定是最后一位的可能是倒数两位三位。。等等

比如0.123...

可能是0.1233333 0.1232323 0.123123123

然后让你计算分母最小的精度与所给的值相等的分数

完全不会啊数学渣渣借鉴别人的博客才大致了解思路

枚举每一种小数点的情况

然后将小数点前的确定精度转化为小数点前的数

然后通过精度差可以计算小数点后所表示的分数然后比较正确精度的分母大小即可

#include <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <cstring>
#include <limits.h>
using namespace std;

int gcd(int a,int b)
{
    if(!a)
        return b;
    return gcd(b%a,a);
}
int main()
{
    char  s[111];
    while(scanf(" %s",s)&&strcmp(s,"0"))
    {
        int num=0;int len=0;int mina=INT_MAX,minb=INT_MAX;
        for(int i=2;s[i]!='.';i++)
        {
            num=num*10+s[i]-'0';
            len++;
        }
        int mid=num;
        for(int i=1;i<=len;i++)
        {
            mid/=10;
            int a=num-mid;
            int b=(int)pow(10.0,len-i)*((int)pow(10,i)-1);
            int j=gcd(a,b);
            if(b/j<minb)
            {
                mina=a/j;
                minb=b/j;
            }
        }
        printf("%d/%d\n",mina,minb);
    }
}