Ducci Sequence

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



这题的意思就是输入n个数，相邻两个数相减的绝对值赋给前一个数，最后一个数与第一个数相减。如此循环下去如果在一千次以内数列的所有数都是0了，就输出ZERO，否则输出LOOP。


代码如下：
  
 

#include <iostream>
using namespace std;
int main()
{
int T;
cin>>T;
while(T--)
{
int n,a[20];
cin>>n;
int flag=0;                     //每次输入完都要把flag归零
for(int i=0;i<n;i++)
cin>>a[i];
for(int j=0;j<1000;j++)
{
int f=a[0];                     //因为先是a[0]与a[1]运算再赋值给a[0],a[0]的值会变，所以要先把a[0]的值赋给f。
int s=0;
for(int i=0;i<n-1;i++)
{
if(a[i]>=a[i+1])
a[i]=a[i]-a[i+1];
else
a[i]=a[i+1]-a[i];
s+=a[i];                     //求数组0到n-2的和。
}
if(f>=a[n-1])
a[n-1]=f-a[n-1];
else
a[n-1]=a[n-1]-f;
s+=a[n-1];               // 再加上a[n-1]的值。
if(s==0)
{
flag=1;
break;
}
}
if(flag)
cout<<"ZERO"<<endl;
else
cout<<"LOOP"<<endl;

}
return 0;
}

第一次写博客，望各位海涵。