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Description
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(3n
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.
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.
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
ZERO LOOP ZERO LOOP
题解：给定数组，依次求前一个减后一个的值的绝对值，如果是最后一个则是最后一个减第一个的值的绝对值，
最多循环1000次，如果出现数组的值全部变为0，则为ZERO，否则为LOOP，所以只求全部为0的情况即可。
#include<iostream>
using namespace std;
int main()
{
int a[20];
int i,j,t,n,sum;
cin>>t;
while(t--)
{
cin>>n;
for(i=0; i<n; i++)
cin>>a[i];
for(i=0; i<1000; i++)
{
sum=0;
int s=a[0];//每循环一次a【0】的值都会变，所以要定义一个变量s
for(j=0; j<n-1; j++)
{
if(a[j]>a[j+1])
a[j]=a[j]-a[j+1];
else
a[j]=a[j+1]-a[j];
sum+=a[j];
}
if(a[n-1]>s)
a[n-1]=a[n-1]-s;
else
a[n-1]=s-a[n-1];
sum+=a[n-1];
if(sum==0)
break;
}
if(sum==0)
cout<<"ZERO"<<endl;
else
cout<<"LOOP"<<endl;
}
return 0;
}
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原文地址：http://www.cnblogs.com/hfc-xx/p/4656271.html
