Uva1594 Ducci Sequence



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

设f(i,j)为第i次做|aj-a(j+1)%n|

由题已知最多1000次之后会出现循环或全部为零,所以只需要判断第f(1001,j)(其中j属于1~n)中有没有大于零的数,若有则是LOOP,否则是ZERO;

#include<stdio.h>#include<algorithm>using namespace std;int a[100];int b[1005][100];int n;void f(int ii){    for (int kk=0;kk<n;kk++)    {        b[0][kk]=a[kk];    }    for (int i=1;i<=ii;i++)    {       for (int j=0;j<n;j++)       {           b[i][j]=abs(b[i-1][j]-b[i-1][(j+1)%n]);       }    }}int main(){    int t;    scanf("%d",&t);    for (int k=0;k<t;k++)    {        scanf("%d",&n);        for (int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        f(1001);        int flag=1;        for (int i=0;i<n;i++)        {            if (b[1001][i]>0) {flag=0; break;}        }        if (flag)        {            printf("ZERO\n");        }        else        {            printf("LOOP\n");        }    }}