zoukankan      html  css  js  c++  java
  • XTU 1236 Fraction

    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;

     
     
     
    看上去很复杂的题,其实是水题,不要被题目吓倒!
    由于分母是1-1000,所以每次将所有的分母枚举一次,选接近的数就可以了。
     
    题意:输入一个小数,输出最接近的分数,必须为最简分数。
     
    附上代码:
     
     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++)  //枚举1-1000的分母
    32         {
    33             j=s*i+0.5;   //求出分子
    34             double f=j*1.0/i; //计算此时分数的结果
    35             double p=fabs(f-s);  //与原来的数进行比较
    36             if(minn>p)
    37             {
    38                 minn=p;
    39                 a=j;
    40                 b=i;
    41             }
    42         }
    43         int r=gcd(a,b); //求最大公约数,化简
    44         printf("%d/%d
    ",a/r,b/r);
    45     }
    46     return 0;
    47 }
  • 相关阅读:
    Java集合框架——Map接口
    深入分析——HashSet是否真的无序?(JDK8)
    你这辈子最引以为傲的是什么?
    全网Star最多(近20k)的Spring Boot开源教程 2019 年要继续更新了!
    Spring Cloud Alibaba与Spring Boot、Spring Cloud之间不得不说的版本关系
    Spring Cloud Alibaba基础教程:Nacos的集群部署
    Spring Cloud Alibaba基础教程:Nacos的数据持久化
    Spring Cloud Alibaba基础教程:Nacos配置的多文件加载与共享配置
    Spring Cloud Alibaba基础教程:Nacos配置的多环境管理
    Spring Cloud Alibaba基础教程:Nacos配置的加载规则详解
  • 原文地址:https://www.cnblogs.com/pshw/p/5572749.html
Copyright © 2011-2022 走看看