Distinct powers
2016-04-10 10:56
316 查看
https://projecteuler.net/problem=29
a ≤ 5 and 2 ≤ b ≤ 5:
22=4, 23=8, 24=16, 25=32
32=9, 33=27, 34=81, 35=243
42=16, 43=64, 44=256, 45=1024
52=25, 53=125, 54=625, 55=3125
If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
How many distinct terms are in the sequence generated by ab for 2 ≤
a ≤ 100 and 2 ≤ b ≤ 100?
方法很简单,算出来放到集合里,最后算一下有多少个,即可。
不然,还要自己去算哪些会重复,忒麻烦。
Distinct powers
Problem 29
Consider all integer combinations of ab for 2 ≤a ≤ 5 and 2 ≤ b ≤ 5:
22=4, 23=8, 24=16, 25=32
32=9, 33=27, 34=81, 35=243
42=16, 43=64, 44=256, 45=1024
52=25, 53=125, 54=625, 55=3125
If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
How many distinct terms are in the sequence generated by ab for 2 ≤
a ≤ 100 and 2 ≤ b ≤ 100?
方法很简单,算出来放到集合里,最后算一下有多少个,即可。
不然,还要自己去算哪些会重复,忒麻烦。
def distinctPowers(): allNumber=set() for i in range(2,101): temp = i for j in range(2,101): temp *= i allNumber.add(temp) return len(allNumber) print(distinctPowers())
相关文章推荐
- 190. Reverse Bits
- Qt编写信息管理系统(2)
- Java集合框架之Map--Hashtable源码分析
- pycharm快捷键及一些常用设置
- 【推荐】可查找下载和上传动漫外挂字幕网站
- HDU 4611 Balls Rearrangement 数学
- 选项卡切换
- Codeforces--658A--Bear and Reverse Radewoosh(模拟)
- shell 之 for 循环
- iOS Md5编码
- 怎样加入cocostudio生成的UI到项目
- Java学习系列——HTTP协议
- 指针与数组
- filter 转换字符编码
- HDOJ-1695 GCD
- 面向对象与面向过程的区别
- Purelayout小结1
- Python文件写操作
- 背包问题:01 完全 多重
- mysql_error() mysql_errno() 错误代码1064