Project Euler Problem 73 (C++和Python代码实现和解析)

Problem 73 : Counting fractions in a range

Consider the fraction, n/d, where n and d are positive integers. If n<d 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?

欧拉项目第73道题 : 1. 计数一个区间的分数

考虑分数n/d，分子n 和 分母d 是正整数。 如果 n<d且n与d互质(n和d的最大公约数是1)，分数n/d被叫做最简真分数。

如果我们按分数的大小的升序列出 d ≤ 8 的最简真分数的集合，则得到
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，在1/3和1/2之间有3个分数。

当d ≤ 12,000时，在排序的最简真分数的集合里，位于1/3和1/2之间的分数有多少？

2. 求解分析

分数n/d是最简真分数(n < d 和HCF(n, d) = 1)，还有 n/d > 1/3 且 n/d < 1/2，那么n > d/3 且 n < d/2，常规方法很容易求解的。

3. C++ 代码实现

C++采用了求解分析的思路，统计了位于1/3和1/2之间的最简真分数的个数。

C++ 代码

#include <iostream>
#include <cassert>
#include <ctime>

using namespace std;

class PE0073
{
private:
    int HCF(int a, int b) 
    {
        int tmp;
        while (b != 0) 
        {
            tmp = a;
            a   = b;
            b   = tmp % b;
        }
        return a;
    }

public:
    int findReducedProperFractions(int max_d);
};

int PE0073::findReducedProperFractions(int max_d)
{
    int numOfFractions = 0;

    for(int d=2; d<=max_d; d++)
    {
        for(int n=d/3; n<=d/2; n++)// n < d
        {
            // n/d > 1/3  and n/d < 1/2 and HCF(n, d) == 1
            if ((d > 2*n) && (d < 3*n) && (1 == HCF(n, d)))
            {
                numOfFractions ++;
            }
        }
    }

    return numOfFractions;
}

int main()
{
    clock_t start = clock();

    PE0073 pe0073;

    assert(3 == pe0073.findReducedProperFractions(8));
    
    int max_d = 12000;
    cout << "For d <= "<<max_d<<", "<< pe0073.findReducedProperFractions(max_d)<<" fractions";
    cout << " lie between 1/3 and 1/2 in the sorted set of reduced proper fractions."<<endl;

    clock_t finish = clock();
    double duration=(double)(finish - start) / CLOCKS_PER_SEC;
    cout << "C/C++ running time: " << duration << " seconds" << endl;

    return 0;
}

4. Python 代码实现

Python和C++一样，采用了求解分析的思路，统计了位于1/3和1/2之间的最简真分数的个数, 但运行时间相比C++太多了。

肯定还有其他更好的求解方法。

Python

import time

def HCF(a, b):
    while b != 0:
        (a, b) = (b, a % b)
    return a

def findReducedProperFractions(max_d):
    numOfFractions = 0

    for d in range(2, max_d+1):
        for n in range(d//3, d//2+1): # n > d/3  and n < d/2
            if (d > 2*n) and (d < 3*n) and (1 == HCF(n, d)):
                numOfFractions += 1

    return numOfFractions

def main():
    start = time.process_time()

    assert 3 == findReducedProperFractions(8)

    max_d = 12000
    total = findReducedProperFractions(max_d)

    print("For d <= %d , %d" %(max_d, total),  "fractions lie between", end=' ')
    print("1/3 and 1/2 in the sorted set of reduced proper fractions")

    end = time.process_time()

    print('PE0073 spent CPU processing time :', end-start)

if  __name__ == '__main__':
    main()