Project Euler Problem 71 (C++和Python代码实现和解析)

Problem 71 : Ordered fractions

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 2/5 is the fraction immediately to the left of 3/7.

By listing the set of reduced proper fractions for d ≤ 1,000,000 in ascending order of size, find the numerator of the fraction immediately to the left of 3/7.

1. 欧拉项目第71道题 : 有序的分数

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，2/5是紧靠3/7的左边的分数。

按大小的升序列出 d ≤ 1,000,000 的最简真分数，找出紧靠3/7的左边的分数的分子。

2. 求解分析

最简真分数n/d： 分子n小于分母d 且分子n和分母d互质。

这个集合有非常多的分数，紧靠3/7的左边的分数有什么特点呢？找出这个特点的话，这道题就能够解答。n/d = 3/7，n = 3d/7, 那么 n’ = (3d-1)/7, 如果n’/d被判断是最简真分数的话，n’/d 就是紧靠3/7的左边的分数，然后，我们从2到max_d枚举，找到分数值最大的那个分数n’/d，就是紧靠3/7的左边的分数。

3. C++ 代码实现

C++代码

#include <iostream>
#include <cassert>

using namespace std;

typedef struct _Fraction 
{
    int n;         // numerator
    int d;         // denominator
    double value;  // n/d
} Fraction;

class PE0071
{
private:
    int HCF(int m, int n) { return (n==0) ? m : HCF(n,  m%n); }

    //  if n<d and HCF(n,d)=1, it is called a reduced proper fraction.
    bool checkReducedProperFraction(int n, int d)
    {
        if (n < d && 1 == HCF(n, d))
        {
            return true;
        }
        return false;
    }
    
public:
    Fraction findReducedProperFractions(int max_d);
};


Fraction PE0071::findReducedProperFractions(int max_d)
{
    Fraction fr;
    Fraction fr_left;

    fr_left.n = 1;
    fr_left.d = max_d;
    fr_left.value = (double)1.0/(double)max_d;

    for(int d=3; d<=max_d; d++)
    {
        int n = (3*d-1)/7;
        if (true == checkReducedProperFraction(n, d))
        {
            fr.n = n;
            fr.d = d;
            fr.value = (double)n/(double)d;

            if (fr.value > fr_left.value)
            {
                fr_left = fr;
            }
        }
    }

    return fr_left;
}

int main()
{
    PE0071 pe0071;

    Fraction fr = pe0071.findReducedProperFractions(8);

    assert(2 == fr.n && 5 == fr.d);

    int max_d = 1000000;

    fr = pe0071.findReducedProperFractions(max_d);

    cout << "For d <= " << max_d << ", " << fr.n << "/" << fr.d;
    cout << " is the fraction immediately to the left of 3/7." << endl;
    
    return 0;
}

4. Python 代码实现

Python 代码

def HCF(a, b):
    """
    Compute the greatest common divisor of a and b. Examples:
    >>> HCF(14, 15)    #co-prime
    1
    """
    while b != 0:
        (a, b) = (b, a % b)
    return a

def checkReducedProperFraction(n, d):
    if n < d and 1 == HCF(n, d):
        return True
    return False

def findReducedProperFractions(max_d):
    fr_left = (1, max_d, 1/max_d)

    for d in range(3, max_d+1):
        n = (3*d-1)/7
        if True == checkReducedProperFraction(n, d):
            fr = (n, d, n/d)

            if fr[2] > fr_left[2]:
                fr_left = fr

    return fr_left

def main():

    fr = findReducedProperFractions(8)
    assert 2 == fr[0] and 5 == fr[1]

    max_d = 10**6
    fr = findReducedProperFractions(max_d)

    print("For d <= %d, %d/%d" %(max_d, fr[0], fr[1]), end='')
    print(" is the fraction immediately to the left of 3/7.")

if  __name__ == '__main__':
    main()