UVA 1594 Ducci Sequence(紫书习题5-2 简单模拟题)

时间:2023-03-09 20:47:36
UVA 1594 Ducci Sequence(紫书习题5-2 简单模拟题)

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

好水的题,,直接莽过去就行了过了,,,或许有不需要循环这么多次的方法。。。

 #include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
int a[];
int n;
bool change() //模拟运算过程
{
int sum = ;
int t = a[];
for(int i = ; i <= n;i++)
{
if(i!=n) a[i] = abs(a[i]-a[i+]);
else a[i] = abs(a[i]-t);
sum+=a[i];
}
if(sum==) return true;
else return false;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i = ;i <= n;i++)
{
scanf("%d",&a[i]);
}
int j;
for( j = ;j <= ; j++)
{
if(change()) break;
}
if(j>) printf("LOOP\n");
else printf("ZERO\n");
}
return ;
}