紫书第五章训练 uva 1594 Ducci Sequence by crq

Description

A Ducci sequence is a sequence of n-tuples of integers. Given an n-tuple of integers (a₁, a₂,^..., a_n), the next n-tuple in the sequence is formed by taking the absolute differences of neighboring integers:

( a₁, a₂, ^..., a_n) (| a₁ - a₂|,| a₂ - a₃|, ^...,| a_n - a₁|)

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:

(8, 11, 2, 7) (3, 9, 5, 1) (6, 4, 4, 2) (2, 0, 2, 4) (2, 2, 2, 2) (0, 0, 0, 0).

The 5-tuple sequence starting with 4,2,0,2,0 enters a loop after 2 steps:

(4, 2, 0, 2, 0) (2, 2, 2, 2, 4) ( 0, 0, 0, 2, 2) (0, 0, 2, 0, 2) (0, 2, 2, 2, 2) (2, 0, 0, 0, 2) (2, 0, 0, 2, 0) (2, 0, 2, 2, 2) (2, 2, 0, 0, 0) (0, 2, 0, 0, 2) (2, 2, 0, 2, 2) (0, 2, 2, 0, 0) (2, 0, 2, 0, 0) (2, 2, 2, 0, 2) (0, 0, 2, 2, 0) (0, 2, 0, 2, 0) (2, 2, 2, 2, 0) ( 0, 0, 0, 2, 2) ^...

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 cases T 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

题意分析：对一个数组不断按照以下过程进行变换（每个数变成后面一个元素与自身相减后取绝对值，最后一个数的下一个是第一个数）：

(a1, a2, · · · , an) → (|a1 − a2|, |a2 − a3|, · · · , |an − a1|)

到最后要么变成全部为0（输出ZERO），要么形成死循环(LOOP)。

解决方法：因为数组元素个数最大才15，直接模拟，数组用vector记录，并用map记录每次变换后数组元素：

map<vector<int>, int> mp;

每次变换后查找map中是否已经存在，若已经存在则输出LOOP，结束。如果变换后结果是全部为0的数组，输出ZERO，结束。

AC源码：

 1 #include <iostream>
 2 #include <string>
 3 #include <vector>
 4 #include <sstream>
 5 #include <map>
 6 using namespace std;
 7 
 8 map<vector<int>, int> mp;
 9 int myabs(int a)
10 {
11     return a>0?a:-a;
12 }
13 
14 bool Zero(vector<int> &vec)
15 {
16     for(int i=0;i<vec.size();i++)
17     {
18         if(vec[i]!=0)
19             return false;
20     }
21     return true;
22 }
23 
24 void Change(vector<int>& vec)
25 {
26     int n = vec.size();
27     int x = vec[0];
28     for(int i=0;i<n-1;i++)
29     {
30         vec[i] = myabs(vec[i+1]-vec[i]);
31     }
32     vec[n-1] = myabs(vec[n-1]-x);
33 }
34 
35 int main()
36 {
37 //    freopen("d:\data1.in","r",stdin);
38     int cas;
39     cin>>cas;
40     while(cas--)
41     {
42         mp.clear();
43         int i, n;
44         vector<int> vec;
45         cin>>n;
46         for(i=0;i<n;i++)
47         {
48             int x;
49             cin>>x;
50             vec.push_back(x);
51         }
52         int flag = 0;
53         while(true)
54         {
55             if(mp[vec])
56             {
57                 flag = 2;//loop
58                 break;
59             }
60             mp[vec] = 1;
61             if(Zero(vec))
62             {
63                 flag = 1;//zero
64                 break;
65             }
66             Change(vec);
67         }
68         if(flag==1)
69             cout<<"ZERO"<<endl;
70         else if(flag==2)
71             cout<<"LOOP"<<endl;
72     } 
73     return 0;
74 }