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.
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 (3 ≤ n ≤ 15), 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.
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
题目大意：给你一个数组，可以对于每个书求出它和下一个数的差的绝对值，得到一个新的数组。重复这个过程，得到的欲裂成为Ducci序列。判断他最终会变成0还是会循环。
思路：每进行一次上述操作就把该数组存入一个set中，然后判断是否出现循环或者全部为0.
#include <bits/stdc++.h>
using namespace std;
vector<int> v,zero;
set< vector<int> > se;
int main(int argc, char const *argv[])
{
int t, n;
scanf("%d",&t);
while(t--){
v.clear();
zero.clear();
se.clear();
scanf("%d",&n);
int temp;
for(int i = 0; i < n; i++){
scanf("%d",&temp);
v.push_back(temp);
zero.push_back(0);
}
se.insert(v);
for(; ;){
if(v == zero){
puts("ZERO");
break;
}
else{
int a = abs(v[0] - v[1]);
for(int i = 1; i < n; i++){
v[i] = abs(v[i] - v[(i + 1) % n]);
}
v[0] = a;
if(se.count(v)){
puts("LOOP");
break;
}
else{
se.insert(v);
}
}
}
}
return 0;
}