XTU1236 Fraction

Posted 2020-10-02 gccbuaa

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 that  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 . 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

题意：把小数化成最接近他的分数，要求最简形式。

分析：枚举分母。每次得到的分子与初始值相比較误差最小的就是答案。

<span style="font-size:18px;">#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
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;
}
</span>