The 3n + 1 problem
Time Limit: 1000MS |
|
Memory Limit: 10000K |
Total Submissions: 39115 |
|
Accepted: 12294 |
Description
Problems in Computer Science are often classified as belonging to a certain class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose classification is not known
for all possible inputs.
Consider the following algorithm:
1. input n
2. print n
3. if n = 1 then STOP
4. if n is odd then n <-- 3n+1
5. else n <-- n/2
6. GOTO 2
Given the input 22, the following sequence of numbers will be printed 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has been verified, however, for all integers n such that
0 < n < 1,000,000 (and, in fact, for many more numbers than this.)
Given an input n, it is possible to determine the number of numbers printed before the 1 is printed. For a given n this is called the cycle-length of n. In the example above, the cycle length of 22 is 16.
For any two numbers i and j you are to determine the maximum cycle length over all numbers between i and j.
Input
The input will consist of a series of pairs of integers i and j, one pair of integers per line. All integers will be less than 10,000 and greater than 0.
You should process all pairs of integers and for each pair determine the maximum cycle length over all integers between and including i and j.
Output
For each pair of input integers i and j you should output i, j, and the maximum cycle length for integers between and including i and j. These three numbers should be separated by at least one space with all three numbers on one
line and with one line of output for each line of input. The integers i and j must appear in the output in the same order in which they appeared in the input and should be followed by the maximum cycle length (on the same line).
Sample Input
1 10
100 200
201 210
900 1000
Sample Output
1 10 20
100 200 125
201 210 89
900 1000 174
Source
这个题,很多人以为是水题。其实大家都没有想想这个题的其他解法
这个题,很多人说事水题的原因是:n太小。所以直接暴力就OK
确实,暴力可以过掉,而且搞不好还是0ms
但是如果n有100w这个时候该怎么暴力?
所以这个时候暴力就是不可取的了
可以观察到,如果要求22的数链长度,其实就是在11的长度上+1,那么要知道11的长度,就在34的长度上面+1,等等等等。。
所以这个题其实可以用记忆优化搜索非常快速的搜索出1到n的所有i的数链长度。
得到长度之后,就可以用线段树求某一个区间的最大值。这一步很显然
看看我的代码吧:
分享到:
相关推荐
北大POJ1207-The 3n + 1 problem 解题报告+AC代码
北大POJ3009-Curling 2.0【DFS+Vector+回溯+剪枝】 解题报告+AC代码
北大POJ3026-Borg Maze【BFS+Prim】 解题报告+AC代码
POJ2186-Popular Cows 【Tarjan+极大强连通分量+缩点】 解题报告+AC代码 http://hi.csdn.net/!s/BGDH68 附:我所有的POJ解题报告链接 . http://blog.csdn.net/lyy289065406/article/details/6642573
北大POJ初级-所有题目AC代码+解题报告
北大POJ3733-Changing Digits【DFS+强剪枝】 解题报告+AC代码
北大POJ3373-Changing Digits【DFS+强剪枝】 解题报告+AC代码
北大POJ3414-Pots 解题报告+AC代码
北大POJ水题整合包 解题报告+AC代码
POJ 1012 约瑟夫问题的数学解法及分析POJ 1012 约瑟夫问题的数学解法及分析POJ 1012 约瑟夫问题的数学解法及分析
自动探测POJ、HDU、SOJ、ZOJ水题,对于有志于刷遍各种水题的ACMer来说非常有用
人们熟悉的四则运算表达式称为中缀表达式,例如(23+34*45/(5+6+7))。在程序设计语言中,可以利用堆栈的方法把中缀表达式转换成保值的后缀表达式(又称逆波兰表示法),并最终变为计算机可以直接执行的指令,得到...
PKU 、POJ ACM/ICPC300多题的代码,还有各种典型问题的分类代码
北大POJ3096-Surprising Strings 解题报告+AC代码
北大POJ1004-Financial Management 解题报告+AC代码
注意,目前POJ还不支持<bits/stdc++.h>(G++、C++都不支持)。HDU部分支持(G++支持,C++不支持)。 其他国外的oj,还有台湾的oj都支持,CF,Topcoder也都支持。 当然,其实这是一个偷懒的写法,但是会降低编译...
很多经典的杭电oj与poj习题的ac代码与详解!全部ac,决对没有错误的代码!
北大POJ初级-简单搜索 解题报告+AC代码
北大POJ1426-Find The Multiple【BFS+同余模】 解题报告+AC代码
北大POJ2676-Sudoku 解题报告+AC代码