Ducci Sequence 解题心得
原题贴上
A Ducci sequence is a sequence of n-tuples of integers. Given an n-tuple of integers (a1, a2, ... , an), the next n-tuple in the sequence is formed by taking the absolute differences of neighboring integers:
Ducci sequences either reach a tuple of zeros or fall into a periodic loop. For example, the 4-tuple sequence starting with 8,11,2,7 takes 5 steps to reach the zeros tuple:
The 5-tuple sequence starting with 4,2,0,2,0 enters a loop after 2 steps:
Given an n-tuple of integers, write a program to decide if the sequence is reaching to a zeros tuple or a periodic loop.
Input
Your program is to read the input from standard input. The input consists of T test cases. The number of test casesT is given in the first line of the input. Each test case starts with a line containing an integer n(3n15), which represents the size of a tuple in the Ducci sequences. In the following line, n integers are given which represents the n-tuple of integers. The range of integers are from 0 to 1,000. You may assume that the maximum number of steps of a Ducci sequence reaching zeros tuple or making a loop does not exceed 1,000.
Output
Your program is to write to standard output. Print exactly one line for each test case. Print `LOOP' if the Ducci sequence falls into a periodic loop, print `ZERO' if the Ducci sequence reaches to a zeros tuple.
The following shows sample input and output for four test cases.
Sample Input
4
4
8 11 2 7
5
4 2 0 2 0
7
0 0 0 0 0 0 0
6
1 2 3 1 2 3
Sample Output
ZERO LOOP ZERO LOOP
分析：
题目让我们求循环出现的时候和全0出现的时候， 其实，全0出现后紧接着就会出现循环，所以我们只需专注于找到循环出现。
这里的循环也就是数字的排列出现重复，也就是发现重复即找到。
想到去重，第一个想到的便是STL中的set可以直接保证每个键是唯一的，然而这里这并不能将重复出现的时候标记出来，达不到解题。接着就很自然的想到“标记法”，即——将每种情形和一个bool值对应，出现过记为true，没出现过记为false。每次新出现一个情形先判断对应的bool值，然后就可以判断重复没有啦。
一种情形与一个值对应，最近正在学STL，然后立马就想到了map。
map<A,B> 应题目要求，A为一个有序对比较适合，所以可以用一个结构体将多个数字组成一个整体，B就很自然的使用bool类型。
可是还有一个问题，map<A,B> 里面的A必须要是定义了<操作符的，因为map判断A的唯一性必须要通过<符号比较，具体的判别机制是（a<b,b<a 有且仅有一个成立 ->a b不相等 //否则a b相等），这就需要我们重载A里的<操作符。
按照判别机制，我如下重载
1 bool operator <(const gro &b) const { 2 for (int i = 0; i < n; i++){ 3 if (g[i] != b.g[i]) return g[i]<b.g[i]; 4 } 5 return false; 6 }
有了这个认识，然后就可以写代码了，如下：
1 #include<iostream>
2 #include<algorithm>
3 #include<map>
4 #include<stdio.h>
5 using namespace std; 6 const int N = 16; 7 bool isloop = 0; 8
9
10 struct gro 11 { 12 int g[N]; 13 int n; 14 void read () 15 { 16 for (int i = 0; i < n; i++) 17 { 18 cin >> g[i]; 19 } 20 } 21 bool iszero(){ 22 for (int i = 0; i < n; i++){ 23 if (g[i] != 0){ 24 return false; 25 } 26 } 27 return true; 28 } 29
30 bool operator <(const gro &b) const { 31 for (int i = 0; i < n; i++){ 32 if (g[i] != b.g[i]) return g[i]<b.g[i]; 33 } 34 return false; 35 } 36
37 void change() 38 { 39 int temp = g[0]; 40 for (int i = 0; i < n - 1; i++) 41 { 42 g[i] = abs(g[i] - g[i + 1]); 43 } 44 g[n - 1] = abs(temp - g[n - 1]); 45 } 46
47 } g1; 48
49
50
51
52
53 map <gro, bool> group; 54
55 int main() 56 { 57 int T; 58 cin >> T; 59 while (T--){ 60 cin >> g1.n; 61 g1.read(); 62 group.clear(); // 因为有多组数据,记得要清空！！！
63 group[g1] = true; 64 for (int k = 0; k < 1000; k++){ 65 g1.change(); 66 if (group[g1] == true){ 67 isloop = true; 68 break; 69 } 70 group[g1] = true; //一定要记得！！！
71 } 72 if (g1.iszero() == true){ 73 puts("ZERO"); 74 } 75 else { 76 puts("LOOP"); 77 } 78 bool isloop = false; 79 } 80
81 return 0; 82 }
骚年，加油奋斗吧，毕竟水题。