python作业第二题 Superpowers

Superpowers

COMP9021, Session 2, 2018

描述

1. 要求输入一行整数，数字之间可以有任意空格，这些整数来代表英雄们的力量值，每人一个力量值，有正有负，英雄力量值可以被翻转数值的相反数，如-1反转为1，7翻转为-7
2. 要求输入一个非负整数 nb_of_switches，最大不超过上一行的英雄数量，该数量代表对英雄的翻转数量
3. 求得一下数值
   1. 求得英雄们翻转次数为nb_of_switches的最大力量值，每个英雄们可以翻转任意次数（包括不翻转，翻转1次，1次以上），如nb_of_switches为3，英雄力量序列为（4，-3，1，2，-7），则可达到的最大力量值为4+3-1+2+7=15
   2. 求得英雄们翻转次数为nb_of_switches的最大力量值，每个英雄们可以最多翻转一次，（完全不翻转、翻转1次），如nb_of_switches为3，英雄力量序列为(4，-3，1，2，-7),则可达到的最大力量值仍未15，nb_of_switches为5，英雄力量序列为（-7, 1, 2, 3, 4），则最大力量值为7-1-2-3-4=-3
   3. 求得英雄们翻转次数为nb_of_switches的最大力量值，被翻转的英雄们必须是连续的nb_of_switches个，例如序列为（4, -3, 1, 2, -7），nb_of_switches为3，则最大值为4 − 3 − 1 − 2 + 7=5
   4. 求得英雄们被翻转后可得的最大力量值，翻转方法为翻转连续的英雄，数量不限（可以不翻转）例如，序列为（ 1 2 3 4 5），最大值为15

def first_power(heros, nb_of_switches):
    for _ in range(nb_of_switches):
        min_ = min(heros)
        pos = heros.index(min_)
        heros[pos] = 0-min_
    s = 0
    for hero in heros:
        s = s + hero
    print('Possibly flipping the power of the same hero many times, the greatest achievable power is %d.' % (s))

def second_power(heros, nb_of_switches):
    switchded_heros = list()
    for _ in range(nb_of_switches):
        min_ = min(heros)
        switchded_heros.append(0-min_)
        heros.remove(min_)
    s = 0
    for hero in heros:
        s = s+hero
    for hero in switchded_heros:
        s = s+hero
    print('Flipping the power of the same hero at most once, the greatest achievable power is %d. ' % (s))

def third_power(heros, nb_of_switches,powerlist,isprint):
    for i in range(len(heros)-nb_of_switches+1):
        current = 0
        s = 0
        for pos in range(len(heros)):
            if (i+current == pos) and (current < nb_of_switches):
                s = s+(0-heros[pos])
                current += 1
            else:
                s = s+heros[pos]
        powerlist.append(s)
    max_ = max(powerlist)
    if isprint:  
        print('Flipping the power of nb_of_flips many consecutive heroes, the greatest achievable power is %d. '%(max_))


def fourth_power(heros):
    powerlist = list()
    for i in range(len(heros)+1):
        third_power(heros,i,powerlist,False)
    max_ = max(powerlist)
    print('Flipping the power of arbitrarily many consecutive heroes, the greatest achievable power is %d. '%(max_))