三个请求，两种做法

运行下面这段程序，填写它打印出来的结果。

DEV = 1000
IRQ = 50

def by_poll(n):
    return n * DEV

def by_irq(n):
    return n * IRQ

def by_dma(n):
    return 2 * IRQ

blocking = 3 * DEV
overlapped = DEV + 2 * IRQ
print(str(blocking) + "/" + str(overlapped) + "/"
      + str(round(blocking / overlapped, 2)))