结果模拟

某棋牌活动的漏洞

33000乐券入局,三人,首局消耗3300每一千得118,等到118×3,兑换3×1000, 则剩余消耗为300

# init token in game
init_token=33000
# money wallet
init_money=0
# for retake 
retake=[1000,118]
#charge 
pay_1=1000
pay_2=pay_1/10
# number
gamer_num=3
# total pay
pay=(pay_1+pay_2)*gamer_num
#cal times
n=0
# loop 
total_pay=0
while init_token>=pay:
    if n >1000:
        break
    else:
        init_token=init_token-pay
        init_money=init_money+pay_1*gamer_num
        n=n+1
        total_pay=pay+total_pay
        init_token=(int(pay/retake[0]))*retake[1]+init_token
        total_pay=total_pay-int(total_pay/retake[0])*retake[0]
        print(int(init_token),init_money,total_pay,n)
次数 账户奖金 扣除兑换后的消耗
1 3000 300
2 6000 600
3 9000 900
4 12000 200
5 15000 500
6 18000 800
7 21000 100
8 24000 400
9 27000 700
10 30000 0
11 33000 300

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转载自www.cnblogs.com/yuvejxke/p/12812402.html