(Problem 73)Counting fractions in a range
Consider the fraction, n/d, where n and d are positive integers. If nd and HCF(n,d)=1, it is called a reduced proper fraction.
If we list the set of reduced proper fractions for d 8 in ascending order of size, we get:
1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
It can be seen that there are 3 fractions between 1/3 and 1/2.
How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions for d 12,000?
题目大意:
考虑分数 n/d, 其中n 和 d 是正整数。如果 nd 并且最大公约数 HCF(n,d)=1, 它被称作一个最简真分数。
如果我们将d 8的最简真分数按照大小的升序列出来,我们得到:
1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
可以看出1/3和1/2之间共有3个分数。
在d 12,000的升序真分数列表中,1/3和1/2之间有多少个分数?
//(Problem 73)Counting fractions in a range
// Completed on Wed, 19 Feb 2014, 16:34
// Language: C11
//
// 版权所有(C)acutus (mail: acutus@126.com)
// 博客地址:http://www.cnblogs.com/acutus/
#include<stdio.h>
#define N 12000 int gcd(int a, int b) //求最大公约数函数
{
int r;
while(b) {
r = a % b;
a = b;
b = r;
}
return a;
} void solve()
{
int a, b, i, j, ans;
ans = ;
for(i = ; i <= N; i++) {
a = i / ; b = i / ;
for(j = a + ; j < b + ; j++) {
if(gcd(i, j) == )
ans++;
}
}
printf("%d\n", ans);
} int main()
{
solve();
return ;
}
Answer:
|
7295372 |
(Problem 73)Counting fractions in a range的更多相关文章
- (Problem 72)Counting fractions
Consider the fraction, n/d, where n and d are positive integers. If nd and HCF(n,d)=1, it is called ...
- (Problem 33)Digit canceling fractions
The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplif ...
- (Problem 35)Circular primes
The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, ...
- (Problem 57)Square root convergents
It is possible to show that the square root of two can be expressed as an infinite continued fractio ...
- (Problem 42)Coded triangle numbers
The nth term of the sequence of triangle numbers is given by, tn = ½n(n+1); so the first ten triangl ...
- (Problem 41)Pandigital prime
We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly o ...
- (Problem 70)Totient permutation
Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number ...
- (Problem 74)Digit factorial chains
The number 145 is well known for the property that the sum of the factorial of its digits is equal t ...
- (Problem 46)Goldbach's other conjecture
It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a ...
随机推荐
- 程序猿的量化交易之路(20)--Cointrader之Assert实体(8)
转载需说明出处:http://blog.csdn.net/minimicall, http://cloudtrade.top 不论什么可交易的都能够称之为Assert,资产.其类代码例如以下: pac ...
- vs2010调试快捷键
VS2010单步调试 1.设置断点 F9设置或者取消断点,如果当前行未设置断点,则F9可以再当前行设置断点,如果已经设置,则为去除断点 2.单步调试 F10不进入函数单步,F11进入函数单步 , ...
- JavaScript之面向对象学习五(JS原生引用类型Array、Object、String等等)的原型对象介绍
1.原型模式的重要性不仅仅体现在创建自定义类型方面,就连所有的原生的引用类型(Obejct.Array.String等等)都在构造函数的原型上定义方法和属性.如下代码可以证明: alert(typeo ...
- WPF利用依赖属性和命令编写自定义控件
以实例讲解(大部分讲解在代码中) 1,新建一个WPF项目,添加一个用户控件之后在用户控件里面添加几个控件用作测试, <UserControl x:Class="SelfControlD ...
- events模块
/** * Created by Administrator on 2016/8/3. */ var http = require("http"); //Node 导入文件系统模块 ...
- C++_基础_继承、多态
内容: (1)子类中的拷贝构造和拷贝赋值 (2)多继承和虚继承 (3)多态的初识 (4)虚析构的特性和使用 (5)多态的底层实现 (6)纯虚函数.抽象类的概念 1.子类中的拷贝构造和拷贝赋值 子类中的 ...
- XLSReadWrite控件简介
2015-10-22 23:57:55 原帖地址:http://www.cnblogs.com/dabiao/archive/2011/07/08/2100609.html XLSReadWrite ...
- windows8 安装IIS 和 添加网站(转)
Internet Information Services(IIS,互联网信息服务),是由微软公司提供的基于运行Microsoft Windows的互联网基本服务.最初是Windows NT版本的可选 ...
- 模块化定义JS,让统一文件夹内相同的变量不冲突
两种方法: 1.(function(){……编写代码……})() //先声明一个函数,声明完后直接调用 2.!function(){……编写代码……}()
- JS计算两个日期相差几天
function Computation(sDate1, sDate2){ var aDate, oDate1, oDate2, iDays aDate = sDate1.split("-& ...