Ducci Sequence
Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %lluDescription
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]; 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; }