Ducci Sequence UVA - 1594
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
HINT
采用思路时使用两个数才存储,其中一个用于计算,另一个用于判断结束,也就是排序找出最大值,如果过最大值为0那就不是循环。另外还要计算判断次数。耗时370ms。
Accepted
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
int m, n,t,i;
cin >> m;
while (m--) {
cin >> i;
n = i;
vector<int>arr,arr2;
while (i--)
{
cin >> t;
arr.push_back(t);
arr2.push_back(t);
}
for (i = 1;;i++)
{
t = arr[0];
for (int j = 0;j < n - 1;j++)
arr[j]= arr2[j] = abs(arr[j] - arr[j + 1]);
arr[n - 1] = arr2[n - 1] = abs(arr[n - 1] - t);
if (i > 1000) {
cout << "LOOP" << endl;
break;
}
sort(arr2.begin(), arr2.end());
if (!arr2[arr2.size() - 1])break;
}
if (i <= 1000)cout << "ZERO" << endl;
}
}
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