Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

A Ducci sequence is a sequence of n-tuples of integers. Given an n-tuple of integers (a₁, a₂, ^... , a_n), 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.

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

---

模擬檢查, 能優化就優化, 不然很容易 TLE

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <map>
using namespace std;
int a[15], stkIdx;
char str[100];
void tran(int n, int st) {
    if(!n && st)
        str[stkIdx++] = '0';
    if(!n)  return;
    tran(n/10, 0);
    str[stkIdx++] = n%10 + '0';
}
void int2str(int n) {
    stkIdx = 0;
    static int i;
    for(i = 0; i < n; i++) {
        tran(a[i], 1);
        str[stkIdx++] = ' ';
    }
    str[stkIdx-1] = '\0';
}
int main() {
    int t, n, i, j;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        int sum = 0;
        for(i = 0; i < n; i++)
            scanf("%d", &a[i]), sum += a[i];
        map<string, bool> r;
        while(sum) {
            int2str(n);
            if(r[str])    break;
            r[str] = true;
            sum = 0;
            for(i = 1, j = a[0]; i < n; i++)
                a[i-1] = abs(a[i-1] - a[i]), sum += a[i-1];
            a[n-1] = abs(a[n-1]-j);
            sum += a[n-1];
        }
        if(sum)
            puts("LOOP");
        else
            puts("ZERO");
    }
    return 0;
}