XTU 1236 Fraction

純轉載，轉載地址：https://www.cnblogs.com/pshw/p/5572749.html

Fraction
Accepted : 168		Submit : 1061
Time Limit : 1000 MS		Memory Limit : 65536 KB

 

Fraction

Problem Description:

Everyone has silly periods, especially for RenShengGe. It's a sunny day, no one knows what happened to RenShengGe, RenShengGe says that he wants to change all decimal fractions between 0 and 1 to fraction. In addtion, he says decimal fractions are too complicate, and set that [Math Processing Error] is much more convient than 0.33333... as an example to support his theory.

So, RenShengGe lists a lot of numbers in textbooks and starts his great work. To his dissapoint, he soon realizes that the denominator of the fraction may be very big which kills the simplicity that support of his theory.

But RenShengGe is famous for his persistence, so he decided to sacrifice some accuracy of fractions. Ok, In his new solution, he confines the denominator in [1,1000] and figure out the least absolute different fractions with the decimal fraction under his restriction. If several fractions satifies the restriction, he chooses the smallest one with simplest formation.

Input

The first line contains a number T(no more than 10000) which represents the number of test cases.

And there followed T lines, each line contains a finite decimal fraction x that satisfies [Math Processing Error] .

Output

For each test case, transform x in RenShengGe's rule.

Sample Input

3
0.9999999999999
0.3333333333333
0.2222222222222

Sample Output

1/1
1/3
2/9

tip

You can use double to save x;

 

 

 

看上去很複雜的題，其實是水題，不要被題目嚇倒！

由於分母是1-1000，所以每次將所有的分母枚舉一次，選接近的數就可以了。

 

題意：輸入一個小數，輸出最接近的分數，必須爲最簡分數。

 

附上代碼：

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 using namespace std;
 6 int gcd(int a,int b)
 7 {
 8     int c,t;
 9     if(a<b)
10     {
11         t=a,a=b,b=t;
12     }
13     while(b)
14     {
15         c=a%b;
16         a=b;
17         b=c;
18     }
19     return a;
20 }
21 int main()
22 {
23     int i,j,T;
24     double s,minn;
25     scanf("%d",&T);
26     while(T--)
27     {
28         scanf("%lf",&s);
29         int a=0,b=1;
30         minn=s;
31         for(i=1; i<=1000; i++)  //枚舉1-1000的分母
32         {
33             j=s*i+0.5;   //求出分子
34             double f=j*1.0/i; //計算此時分數的結果
35             double p=fabs(f-s);  //與原來的數進行比較
36             if(minn>p)
37             {
38                 minn=p;
39                 a=j;
40                 b=i;
41             }
42         }
43         int r=gcd(a,b); //求最大公約數，化簡
44         printf("%d/%d\n",a/r,b/r);
45     }
46     return 0;
47 }