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XTU1236 Fraction

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XTU1236 Fraction

Fraction

Accepted : 124 Submit : 806 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 thatis 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.

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;

Source

XTU OnlineJudge 題意：把小數化成最接近他的分數，要求最簡形式。分析：枚舉分母，每次得到的分子與初始值相比較誤差最小的就是答案。

#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define lson (i<<1)
#define rson ((i<<1)|1)
#define MAXN 100010

int main()
{
    int T;
    double s,minx;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf",&s);
        minx = s;
        int a=0,b=1;
        for(int i=1; i<=1000; i++)
        {
            int j = i*s+0.5;
            double f = j*1.0/i;
            double p = fabs(f-s);
            if(minx > p)
            {
                minx = p;
                a = j;
                b = i;
            }
        }
        int r = __gcd(a, b);
        printf("%d/%d\n",a/r,b/r);
    }
    return 0;
}