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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:
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:
The 5-tuple sequence starting with 4,2,0,2,0 enters a loop after 2 steps:
Given an n-tuple of integers, write a program to decide if the sequence is reaching to a zeros tuple or a periodic loop.
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.
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.
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
ZERO LOOP ZERO LOOP
题意:给你一个数组,相邻数相加(最后一个数应该加第一个数),然后又可以得到一个新的数组,反复进行这样的步骤,结果会有两种,数组每个成员都是零,或者在这样的过程中出现了周期,即出现了和以前已经出现过的相同的项。
最后要求判断是哪一种情况
思路:
如果按照一般的解题思想,每得到一个新的数组都去判断是zero,还是loop。zero还好说,如果判断loop那就要每次遍历一遍数组,程序肯定会超时。、
所以~~~
反正结果只有两种可能,不是loop,就是zero,只判断zero就好了
代码:
#include"iostream"
#include"cstring"
using namespace std;
const int maxn=1010;
const int maxm=16;
int len;
int loop=0,zero=0;
int a[maxn][maxm];
int swift(int number)
{
return (number<0)?-number:number;
}
void Init()
{
cin>>len;
for(int i=0;i<len;i++)
cin>>a[0][i];
}
void Work()
{
int flag=0;
for(int j=1;j<maxn;j++)
{
for(int i=0;i<len;i++)
{
if(i==len-1)
a[j][i]=swift(a[j-1][i]-a[j-1][0]);
else
a[j][i]=swift(a[j-1][i]-a[j-1][i+1]);
}
for(int h=0;h<len;h++)
if(a[j][h]!=0) {flag=1;break;}
if(flag==0) {zero=1;return;}
else flag=0;
}
}
int main()
{
int n;
cin>>n;
while(n--)
{
Init();
Work();
if(zero)
{
cout<<"ZERO"<<endl;
zero=0;
}
else
cout<<"LOOP"<<endl;
}
return 0;
}
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原文地址:http://www.cnblogs.com/zsyacm666666/p/4652545.html