原题传送门

1. 题目描述

2. Solution

1、思路
设要拆解的真分数为\(\frac{A}{B}\)，设 B除以A的商为C，余数为D。即: \(B=A \cdot C+D\)。等式两边同时除以A，可得
\(\frac{B}{A} = C + \frac{D}{A} < C + 1(\because D < A)\) ，
易知C + 1是大于\(\frac{B}{A}\)的最小分数。将分式倒置有
\(\frac{A}{B} > \frac{1}{C+1}\)
即为真分数\(\frac{A}{B}\)包含的最大埃及分数。设\(E=C+1\)，则有
\(\frac{A}{B} - \frac{1}{E} = \frac{AE-B}{BE}\)
继续上面的算法，直到A=1。
总结算法:

1. 取\(g = gcd(A, B), A = A / g, B = B / g，除掉最大公约数\)

2. \(C = \left[ \frac{B}{A} \right] 向上取整, E = C + 1，导出 \frac{1}{E}\)

3. \(A =AE-B, B = BE，迭代计算，直到A=1\)

   2、实现
   最简单的实现自然是递归

import sys

if sys.platform != "linux":
    file_in = open("input/HJ82.txt")
    sys.stdin = file_in


def gcd(a, b):
    return a if b == 0 else gcd(b, a % b)


def solve(A, B):
    if A == 1:
        print(f"1/{B}")
        return
    g = gcd(A, B)
    A = A // g
    B = B // g
    C = B // A
    E = C + 1
    print(f"1/{E}", end="+")
    solve(A * E - B, B * E)


for line in sys.stdin:
    A, B = list(map(int, line.strip().split("/")))
    solve(A, B)

标签：真分数,frac,gcd,sys,牛客,solve,file,机试,HJ82
来源： https://www.cnblogs.com/junstat/p/16177295.html