Huas Summer training#1 B

Topic:

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 are formed by taking the absolute differences of neighboring integers:

( a 1, 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 and 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 are reaching to a zeros tuple or a per Iodic Loop.

Input

Your program was to read the input from standard input. The input consists of T test Cases. The number of test cases T is given on the first line of the input. Each test case starts with a line containing an integer n(3n), which represents the size of a Tuple in the Ducci sequences. In the following line, n integers is given which represents the n-tuple of integers. The range of integers is from 0 to 1,000. Assume that the maximum number of steps of a ducci sequence reaching zeros tuples or making a loop does not exceed A.

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 Reache s to a zeros tuple.

The following shows sample input and output for four test cases.

The main topic: give you n number, with ( a1, a2,..., aN) (| a1- a2|,| a2- a3|, ..., | aN- a1|) To find it in 1000 calculations is the output loop no output zero; a problem-solving idea: To store these numbers in a vector array for comparison; code:

1#include <iostream>2#include <math.h>3#include <vector>4 using namespacestd;5 Const  intL= ++Ten;6 7 intMain ()8 {9vector<int>A[l];Tenvector<int>b; One     intt,n,i,t,j,d,p; ACin>>T; -      while(t--) -     { theCin>>N; -         if(n>=3&&n<= the) -         { -D=0; + b.resize (n); -              for(i=0; i<l;i++) + a[i].resize (n); A              for(i=0; i<n;i++) at             { -scanf"%d", &a[0][i]); -b[i]=0; -                 if(a[0][i]<0|| a[0][i]> +) -scanf"%d", &a[0][i]); -             } in              for(i=0;i< +; i++)  -             { tot=a[i][0]; +                  for(j=0; j<n;j++)  -                 {         the                     if(j== (n1)) *a[i+1][j]=abs (a[i][n-1]-t);  $                     Else Panax Notoginsenga[i+1][j]=abs (a[i][j]-a[i][j+1]); -                 } the                 if(a[i+1]==b) +                 { Aprintf"zero\n"); theD=i; +                      Break; -                 } $             } $             if(d==0) -printf"loop\n");  -         }                 the     } -     return 0;

Huas Summer training#1 B