利用条件运算符的嵌套来完成此题:学习成绩>=90分的同学用A表示,60-80分之间的用B表示,60分以下的用C表示。思路:(a>b)?a:b这是条件运算符的基本例子。score = int(input('input score:'))if score >= 90:grade = 'A'______________grade = 'B'elif score >= 70:grade = 'C'elif score >= 60:grade = 'D'else:grade = 'E'print ('%d belongs to %s' % (score,grade))
A. elif score > 80:
B. elif score < 80:
C. elif score <= 80:
D. elif score >= 80:
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求s=a+aa+aaa+aaaa+aa...a的值,其中a是一个数字。例如2+22+222+222...2(此时共有5个数相加),几个数相加有键盘控制。需求:求s=a+aa+aaa+aaaa+aa...a的值,其中a是一个数字。思路:先计算出每一项的值。Tn = 0Sn = []n = int(input('n = :'))a = int(input('a = :'))for count in range(n):#先循环计算每一项的值Tn = Tn + a_______________Sn.append(Tn)print TnSn = reduce(lambda x,y : x + y,Sn)#在计算和print( Sn)
A. a = a // 10
B. a = a + 10
C. a = a * *10
D. a = a * 10
一个整数,它加上100后是一个完全平方数,再加上168又是一个完全平方数,请问该书是多少?import mathfor i in range(10000):x = int(math.sqrt(i + 100))y = int(math.sqrt(i + 168))if ____________________________________:print(i)
A. (x * x == i + 100) and (y * y == i + 168)
B. (x * x == i + 100) && (y * y == i + 168)
C. (x *2 == i + 100) and (y **2 == i + 168)
D. (x * x = i + 100) and (y * y = i + 168)
输入某年某年某月某日,判断这一天是这一年的第几天?思路:以3月5日为例,应该先把前两个月的加起来,然后再加上5天即本年的第几天,特殊情况,闰年且输入月份大于3时需考虑多加一天:year = int(input('year:\n'))month = int(input('month:\n'))day = int(input('day:\n'))months = [0,31,59,90,120,151,181,212,243,273,304,334]if 0 <= month <= 12:#判断输入的月份sum = months[month - 1]else:print('data error')sum += dayleap = 0if ________________________________:#判断是平年还是润年leap = 1if (leap == 1) and (month > 2):sum += 1print ('it is the %dth day.' % sum)
A. (year % 400 == 0) and ((year % 4 == 0) and (year % 100 != 0))
B. (year % 400 == 0) and ((year % 4 == 0) or (year % 100 != 0))
C. (year % 400 == 0) or ((year % 4 == 0) and (year % 100 != 0))
D. (year % 400 = 0) or ((year % 4 = 0) and (year % 100 != 0))
输出9*9乘法口诀表。思路:分行与列考虑,共9行9列,i控制行,j控制列。'''for i in range(1,10):______________________:result = i * jprint ('%d * %d = % - 3d'% (i,j,result)) #循环的两个数相乘print ('')#两个循环的间隔
A. for j in range(1,10)
B. for j in range(1,9)
C. for j in range(9)
D. for j in range(10)