枚举——完美立方

文章作者：Tyan
博客：noahsnail.com | CSDN | 简书

1. 枚举

枚举是基于逐个尝试答案的一种问题求解策略。

2. 完美立方

形如$a^3 = b^3 + c^3 + d^3$的等式被称为完美立方等式。例如$12^3 = 6^3 + 8^3 + 10^3$

问题：编写程序，对任给的正整数N（N<=100），寻找所有的四元组(a, b, c, d)，使得$a^3 = b^3 + c^3 + d^3$，其中a，b，c，d大于1，小于等于N，且b<=c<=d。

输入：一个正整数N（N<=100）。
输出：每行输出一个完美立方。输出格式为Cube = a，Triple = (b, c, d)。

求解：

备注：判断条件边界很重要

方法一：

#!/usr/bin/env python
# _*_ coding: utf-8 _*_


import sys
import math

# n = sys.argv[1]
n = 24

i = 0
for a in xrange(2, n + 1):
    for b in xrange(2, n):
        for c in xrange(b, n):
            for d in xrange(c, n):
                i += 1
                if math.pow(a, 3) == math.pow(b, 3) + math.pow(c, 3) + math.pow(d, 3):
                    print 'Cube = %d, Triple = (%d, %d, %d)' % (a, b, c, d)
print '%d iterations.' % i

输出

Cube = 6, Triple = (3, 4, 5)
Cube = 12, Triple = (6, 8, 10)
Cube = 18, Triple = (2, 12, 16)
Cube = 18, Triple = (9, 12, 15)
Cube = 19, Triple = (3, 10, 18)
Cube = 20, Triple = (7, 14, 17)
Cube = 24, Triple = (12, 16, 20)
46552 iterations.

方法二

#!/usr/bin/env python
# _*_ coding: utf-8 _*_


import sys
import math

# n = sys.argv[1]
n = 24

i = 0
for a in xrange(2, n + 1):
    for b in xrange(2, a):
        for c in xrange(b, a):
            for d in xrange(c, a):
                i += 1
                if math.pow(a, 3) == math.pow(b, 3) + math.pow(c, 3) + math.pow(d, 3):
                    print 'Cube = %d, Triple = (%d, %d, %d)' % (a, b, c, d)
print '%d iterations.' % i

输出

Cube = 6, Triple = (3, 4, 5)
Cube = 12, Triple = (6, 8, 10)
Cube = 18, Triple = (2, 12, 16)
Cube = 18, Triple = (9, 12, 15)
Cube = 19, Triple = (3, 10, 18)
Cube = 20, Triple = (7, 14, 17)
Cube = 24, Triple = (12, 16, 20)
12650 iterations.

从上面可以看出枚举的边界不同，效率会差将近三倍。

Python源码地址：https://github.com/SnailTyan/programming-and-algorithms/blob/master/perfect_cubes.py，记得给个star。

C++源码地址（已在POJ上Accepted）：https://github.com/SnailTyan/programming-and-algorithms/blob/master/perfect_cubes.cpp，记得给个star。