如果某个数 K 的平方乘以 N 以后，结果的末尾几位数等于 K，那么就称这个数为“N-自守数”。例如 3×92²=25392，而 25392的末尾两位正好是 92，所以 92 是一个 3-自守数。

本题就请你编写程序判断一个给定的数字是否关于某个 N 是 N-自守数。

输入格式：

输入在第一行中给出正整数 M（≤20），随后一行给出 M 个待检测的、不超过 1000 的正整数。

输出格式：

对每个需要检测的数字，如果它是 N-自守数就在一行中输出最小的 N 和 NK² 的值，以一个空格隔开；否则输出 No。注意题目保证 N<10。

输入样例：

3
92 5 233

输出样例：

3 25392
1 25
No

编程语言：Python

解题思想：

1.穷举N的每种情况

2.利用字符串切片的操作判断是否为自守数

时间复杂度：O(N^2)

代码如下：

NON = int(input())
 
Number = list(map(int, input().split()))  # 将一行的输入转为列表
 
for i in range(NON):  # 遍历该列表
    K = Number[i]
    Judge = False  # 标记该数是否为自守数，默认为否
    for j in range(10):  # 穷举N的每种情况
        N = j
        for k in range(len(str(K * K * N)), -1, -1):  # 遍历字符串的长度，例如265962，顺序为6、5、4···
            if str(K) == str(K * K * N)[k:]:  # 例如字符串为26532，[k:]的作用在于取字符串的后5-k位
                print(N, K * K * N)
                Judge = True
                break
        if Judge:
            break
    if not Judge:
        print("No")

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PAT乙级:1091 N-自守数