文章作者：Tyan
博客：noahsnail.com | CSDN | | 云+社区
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$a^3 = b^3 + c^3 + d3$的等式被称为完美立方等式。例如$123 = 6^3 + 8^3 + 10^3$
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。
参考资料
- 程序设计与算法（二）算法基础