子曰：“三人行，必有我师焉。择其善者而从之，其不善者而改之。”

本题给定甲、乙、丙三个人的能力值关系为：甲的能力值确定是 2 位正整数；把甲的能力值的 2 个数字调换位置就是乙的能力值；甲乙两人能力差是丙的能力值的 X 倍；乙的能力值是丙的 Y 倍。请你指出谁比你强应“从之”，谁比你弱应“改之”。

输入格式：

输入在一行中给出三个数，依次为：M（你自己的能力值）、X 和 Y。三个数字均为不超过 1000 的正整数。

输出格式：

在一行中首先输出甲的能力值，随后依次输出甲、乙、丙三人与你的关系：如果其比你强，输出 Cong；平等则输出 Ping；比你弱则输出 Gai。其间以 1 个空格分隔，行首尾不得有多余空格。

注意：如果解不唯一，则以甲的最大解为准进行判断；如果解不存在，则输出 No Solution。

输入样例 1：

48 3 7

输出样例 1：

48 Ping Cong Gai

输入样例 2：

48 11 6

输出样例 2：

No Solution

编程语言：Python

解题思想：

1.在10-100中枚举甲的能力值

2.简单的数学变换和判断

时间复杂度：O(N)

代码如下：

M, X, Y = map(int, input().split())
P1, P2, P3 = 0, 0, 0  # 存放甲、乙、丙的能力值
Result = []  # 存放甲的最大解时的甲、乙、丙的能力值
 
for i in range(10, 100):
    P1 = i
    P2 = (P1 % 10) * 10 + (P1 // 10)  # 将甲的能力值转置，得到乙的能力值
    if abs(P1 - P2) / X == P2 / Y:  # 存在符合条件的丙的能力值
        P3 = abs(P1 - P2) / X
        Result = [P1, P2, P3]  # 覆盖
 
if len(Result) != 0:  # 有解
    print(Result[0], end=' ')
    for i in range(3):
        if M < Result[i]:
            Result[i] = "Cong"
        elif M == Result[i]:
            Result[i] = "Ping"
        elif M > Result[i]:
            Result[i] = 'Gai'
    print(Result[0], Result[1], Result[2])
else:  # 无解
    print("No Solution")

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