1053 Python implementation housing vacancy rate

tags: PAT Exercises Python language programming

1053 housing vacancy rate (20 points)

Without disturb residents, a method of statistical housing vacancy rate is judged by a continuous variation of electricity according to household. Determined as follows:

During the observation period, if there is more than half of the day electricity is below a given threshold value E, the housing is "vacant";
When the observation period exceeds a given threshold value D days, and on a condition satisfied, the housing is "empty."

Now given a residential area of household electricity consumption data, you statistics "vacant" ratios and "vacancy" rate, housing that is more than two states as a percentage of total units of residential housing.

Input formats:

Input of the first row is given a positive integer N (≤1000), total units of residential housing; positive real number to e, the low battery threshold; positive integer D, the observation period threshold. Then N rows, each housing a given power consumption data in the following format:

K E1 E2 ... EK

Where K is the number of days of observation, E i is the i-day consumption.

Output formats:

The percentage ratio of the output values in a row "vacant" and "empty" ratio, with a space therebetween, a retained after the decimal point.

Sample input:

5 0.5 10
6 0.3 0.4 0.5 0.2 0.8 0.6
10 0.0 0.1 0.2 0.3 0.0 0.8 0.6 0.7 0.0 0.5
5 0.4 0.3 0.5 0.1 0.7
11 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1
11 2 2 2 1 1 0.1 1 0.1 0.1 0.1 0.1

Sample output:

40.0% 20.0%

(Sample explained: the second and third family is "vacant", the first four as "vacant", other users are not empty.)

Analysis: statistics "vacant" and "empty" the number, the conversion ratio percentile can, pay attention, first determine when the vacant vacant judge again, determined to be vacant, then it is not vacant, the vacancy may be in subtracting a

program:

a = input().split()
m = 0
n = 0
for i in range(int(a[0])):
    b = input().split()
    x = 0
    for j in range(1,len(b)):
        if float(b[j])<float(a[1]):
            x+=1
    if x>int(b[0])//2:
        m+=1
        if int(b[0])>int(a[2]):
            n+=1
            m-=1
m = m/int(a[0])*100
n = n/int(a[0])*100
m = '%.1f'%m
n = '%.1f'%n
print("%s%c%c%s%c"%(m,'%',' ',n,'%'),end="")

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