ZOJ--003Crashing Balloon

题目描述

Crashing Balloon：问题可以简化为判断两个数是否一定有相同公因子。且每个数的因子集合中不能有相同因子，且每个因子都属于集合{1,2,…,100}.

题目详细描述请参考ZOJ

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解题

思路：如果数字冲突，小胜；反之，大胜。

代码如下：

import sys

factors = []


def factoring(buf, num):
    '''
    :param buffer: 缓冲备份列表
    :param num: 待分解整数
    :return:用factors列表返回分解结果,要求因子各不相同
    '''
    global factors
    i = 2
    while i <= num:
        if i > 100:
            break
        if num % i == 0 and i not in buf:
            buf.append(i)
            factor = buf[:]
            temp = int(num / i)
            if temp not in factor and temp < 101:
                if temp != 1:
                    factor.append(temp)
                factor.sort()
                if factor not in factors:
                    factors.append(factor)
            factoring(buf, temp)
            buf.pop()
        i += 1
    return factors


def winner(num1, num2):
    '''
    :param num1:
    :param num2:
    :return: 获胜者
    '''
    global factors
    if num1 == num2:
        return num1
    if num1 < num2:
        num1, num2 = num2, num1
    factors = []
    factors1 = factoring([], num1)
    factors = []
    # 非常关键
    if num2 == 1:
        factors2 = [[1]]
    else:
        factors2 = factoring([], num2)
    # print('num1:', factors1)
    # print('num2:', factors2)
    if len(factors2) == 0:
        # num2假   num1获胜
        return num1
    elif len(factors1) == 0:
        # num1假 num2真  num2获胜
        return num2
    else:
        # num1、num2皆真
        for factor1 in factors1:
            for factor2 in factors2:
                if len(set(factor1).intersection(set(factor2))) == 0:
                    # 存在不冲突  num1获胜
                    return num1
        # 完全冲突  num2获胜
        return num2


if __name__ == '__main__':
    for line in sys.stdin:
        s = []
        s = line.split()
        print(winner(int(s[0]), int(s[1])))

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