紫书第五章训练 uva 1594 Ducci Sequence by crq
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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.
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
题意分析:对一个数组不断按照以下过程进行变换(每个数变成后面一个元素与自身相减后取绝对值,最后一个数的下一个是第一个数):
(a1, a2, · · · , an) → (|a1 ? a2|, |a2 ? a3|, · · · , |an ? a1|)
到最后要么变成全部为0(输出ZERO), 要么形成死循环(LOOP)。
解决方法:因为数组元素个数最大才15,直接模拟,数组用vector记录,并用map记录每次变换后数组元素:
map<vector<int>, int> mp;
每次变换后查找map中是否已经存在,若已经存在则输出LOOP,结束。如果变换后结果是全部为0的数组,输出ZERO,结束。
AC源码:
1 #include <iostream> 2 #include <string> 3 #include <vector> 4 #include <sstream> 5 #include <map> 6 using namespace std; 7 8 map<vector<int>, int> mp; 9 int myabs(int a) 10 { 11 return a>0?a:-a; 12 } 13 14 bool Zero(vector<int> &vec) 15 { 16 for(int i=0;i<vec.size();i++) 17 { 18 if(vec[i]!=0) 19 return false; 20 } 21 return true; 22 } 23 24 void Change(vector<int>& vec) 25 { 26 int n = vec.size(); 27 int x = vec[0]; 28 for(int i=0;i<n-1;i++) 29 { 30 vec[i] = myabs(vec[i+1]-vec[i]); 31 } 32 vec[n-1] = myabs(vec[n-1]-x); 33 } 34 35 int main() 36 { 37 // freopen("d:\\data1.in","r",stdin); 38 int cas; 39 cin>>cas; 40 while(cas--) 41 { 42 mp.clear(); 43 int i, n; 44 vector<int> vec; 45 cin>>n; 46 for(i=0;i<n;i++) 47 { 48 int x; 49 cin>>x; 50 vec.push_back(x); 51 } 52 int flag = 0; 53 while(true) 54 { 55 if(mp[vec]) 56 { 57 flag = 2;//loop 58 break; 59 } 60 mp[vec] = 1; 61 if(Zero(vec)) 62 { 63 flag = 1;//zero 64 break; 65 } 66 Change(vec); 67 } 68 if(flag==1) 69 cout<<"ZERO"<<endl; 70 else if(flag==2) 71 cout<<"LOOP"<<endl; 72 } 73 return 0; 74 }
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