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。。。
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 using namespace std; 5 int a[20]; 6 int n; 7 bool change() //模拟运算过程 8 { 9 int sum = 0; 10 int t = a[1]; 11 for(int i = 1; i <= n;i++) 12 { 13 if(i!=n) a[i] = abs(a[i]-a[i+1]); 14 else a[i] = abs(a[i]-t); 15 sum+=a[i]; 16 } 17 if(sum==0) return true; 18 else return false; 19 } 20 int main() 21 { 22 int T; 23 scanf("%d",&T); 24 while(T--) 25 { 26 scanf("%d",&n); 27 for(int i = 1;i <= n;i++) 28 { 29 scanf("%d",&a[i]); 30 } 31 int j; 32 for( j = 1;j <= 1000 ; j++) 33 { 34 if(change()) break; 35 } 36 if(j>1000) printf("LOOP "); 37 else printf("ZERO "); 38 } 39 return 0; 40 }