题目：
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:
( a1, a2, ... , an) 
(| a1 - a2|,| a2 - a3|, ... ,| an - 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.
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(3
n15), 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.
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.
题目大意：给你N个数，用( a1, a2, ... , an) 
(| a1 - a2|,| a2 - a3|, ... ,| an - a1|)在1000次计算里找它的是否循环 是输出LOOP 否输出ZERO；
解题思路：用vector数组存储这些数 在进行比较；
代码：
1 #include<iostream>
2 #include<math.h>
3 #include<vector>
4 using namespace std;
5 const int l=1000+10;
6
7 int main()
8 {
9 vector<int> a[l];
10 vector<int> b;
11 int T,n,i,t,j,d,p;
12 cin>>T;
13 while(T--)
14 {
15 cin>>n;
16 if(n>=3&&n<=15)
17 {
18 d=0;
19 b.resize(n);
20 for(i=0;i<l;i++)
21 a[i].resize(n);
22 for( i=0;i<n;i++)
23 {
24 scanf("%d",&a[0][i]);
25 b[i]=0;
26 if(a[0][i]<0||a[0][i]>1000)
27 scanf("%d",&a[0][i]);
28 }
29 for(i=0;i<1000;i++)
30 {
31 t=a[i][0];
32 for(j=0;j<n;j++)
33 {
34 if(j==(n-1))
35 a[i+1][j]=abs(a[i][n-1]-t);
36 else
37 a[i+1][j]=abs(a[i][j]-a[i][j+1]);
38 }
39 if(a[i+1]==b)
40 {
41 printf("ZERO\n");
42 d=i;
43 break;
44 }
45 }
46 if(d==0)
47 printf("LOOP\n");
48 }
49 }
50 return 0;