A Ducci sequence is a sequence of n-tuples of integers. Given an n-tuple of integers (a1a2, ... an), the next n-tuple in the sequence is formed by taking the absolute differences of neighboring integers:

a1a2... 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(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:给你一个数组,相邻数相加(最后一个数应该加第一个数),然后又可以得到一个新的数组,反复进行这样的步骤,结果会有两种,数组每个成员都是零,或者在这样的过程中出现了周期,即出现了和以前已经出现过的相同的项。
最后要求判断是哪一种情况 思路:
如果按照一般的解题思想,每得到一个新的数组都去判断是zero,还是loop。zero还好说,如果判断loop那就要每次遍历一遍数组,程序肯定会超时。、
所以~~~
反正结果只有两种可能,不是loop,就是zero,只判断zero就好了 代码:
#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=;
int a[][maxn];
int n;
int iszero; void Init()
{
cin>>n;
for(int i=;i<n;i++)
cin>>a[][i];
} bool judge()
{
iszero=;
int now,form;
for(int i=;i<=;i++)
{
if(i%==) now=,form=;
else now=,form=;
for(int j=;j<n;j++)
{
if(j==n-)
{
a[now][j]=a[form][j]+a[form][];
}
else
a[now][j]=a[form][j]+a[form][j+];
}
int flag=;
for(int k=;k<n;k++)
if(a[now][k]!=) {flag=;break;}
if(!flag) {iszero=;break;}
}
if(iszero) return true;
else return false;
} int main()
{
int T;
cin>>T;
while(T--)
{
Init();
if(judge()) cout<<"ZERO"<<endl;
else cout<<"LOOP"<<endl;
}
return ;
}
 

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