流程控制练习题
使用尽可能多的方法实现list去重
方法1：
a=[1,1,2,2,3,3,4,5]
print set(a)
方法2：
a=[1,1,2,2,3,3,4,5]
b=[]
for i in a:
if i in b:
continue
else:
b.append(i)
print b
方法3:
a=[1,1,2,2,3,3,4,5]
d={}
for i in a:
d[i]=""
print d.keys()
成绩等级判断
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，60-89分之间的用B表示，60分以下的用C表示
while 1:
grade=raw_input("please input your grade:")
if grade==".":
break
else:
grade=int(grade)
if grade>=90:
print "your level is A"
elif grade >=60 and grade <90:
print "your level is B"
else:
print "your level is C"
实现数学中多项式求和公式的打印
a6x^6 + a5x^5 + a4x^4 + a3x^3 + a2x^2 + a1x^1 + a0
exp=""
for i in range(6,0,-1):
exp+=('a%s^x%s+' %(i,i))
exp+='a0'
print exp
统计名字列表中，各名字的首字母在名字列表中出现的次数
name_list=["jane","fosterwu","janet","jordan","flore",'michel']
d={}
for name in name_list:
if d.has_key(name[0]):
d[name[0]]+=1
else:
d[name[0]] = 1
for letter,times in d.items():
print "%s occurs %s times" %(letter,times)
输入三个数，判断是否能构成三角形
能构成三角形三边关系：
三边都大于零
两边之和大于第三边，两边之差小于第三边
a=int(raw_input("please input side length:"))
b=int(raw_input("please input side length:"))
c=int(raw_input("please input side length:"))
if a>0 and b>0 and c>0 and a+b>c and a+c>b and b+c>a and a-b<c and a-c<b and b-c<a:
print "it is a triangle"
else:
print "it is not a triangle"
实现字典的fromkeys方法
def from_keys(lista,value):
d={}
for i in lista:
d[i]=value
return d
print from_keys(['a',"b","c"],10)
键盘读入一字符串，逆序输出：
line=raw_input("input a sentence:")
print line[::-1]
读入一个整数n，输出n的阶乘
def factorial(n):
result=1
for i in range(1,n+1):
result*=i
return result
print factorial(5)
打印1/2, 1/3, 1/4,….1/10
result=[]
for i in range(2,11):
result.append("1/"+str(i))
print ",".join(result)
写一个函数实现一个数学公式
import math
def area(radius):
return math.pi*radius**2
print area(10)
输入数字a，n，如a，4，则打印a+aa+aaa+aaaa之和
def print_exp(letter,times):
result=[]
for i in range(1,times+1):
result.append(letter*i)
return result
print "+".join(print_exp("a",4))
求100个随机数之和，随机数要求为0—9的整数
import random
result=0
for i in range(100):
result+=random.randint(0,9)
print result
要求在屏幕上分别显求1到100之间奇数之和与偶数之和
odd_number_sum=0
oven_number_sum=0
for i in range(1,101):
if i%2==0:
oven_number_sum+=i
else:
odd_number_sum+=i
print odd_number_sum
print oven_number_sum
输入10个数，并显示最大的数与最小的数
biggest_number=None
least_number=None
for i in range(5):
number=float(raw_input("please input a number:"))
if i ==0:
biggest_number,least_number=i,i
if number>biggest_number:
biggest_number=number
if number<least_number:
least_number=number
print biggest_number
print least_number
给一个不多于5位的正整数，要求：一、求它是几位数，二、逆序打印出各位数字。
number=raw_input("please input a number:")
print "number is %s length" %(len(number))
print str(number)[::-1]
求1000以内的所有水仙花数
for i in range(100,1000):
length=len(str(i))
if int(str(i)[0])**length+int(str(i)[1])**length+int(str(i)[2])**length==i:
print i
编程求s=1!+2!+3!+…..+n!
def factorial(n):
result=1
for i in range(1,n+1):
result*=i
return result
def compute_factorial_sum(n):
sum=0
for i in range(1,n+1):
sum+=factorial(i)
return sum
print compute_factorial_sum(3)
钞票换硬币
result=[]
for i in range(0,101):
for j in range(0,51):
for m in range(0,21):
if i+j*2+m*5==100:
result.append("%s,%s,%s" %(i,j,m))
print len(result)
print result
自己实现在一句话中查找某个单词的算法，存在返回索引号，否则返回False
sentence="i am a boy!"
search_word="i"
def find_word(word,target_sentence):
word_length=len(word)
for i in range(len(target_sentence)):
if sentence[i:i+word_length]==word:
return i
return False
print find_word(search_word,sentence)
读入一个十进制整数，实现十进制转二进制算法将其转成二进制数要求：不能使用现成进制转换函数，自己写代码实现
请见讲义
将一个矩阵转置
a=[[1,2,3],[4,5,6]]
def covert_array(array):
b=[]
for i in range(len(a[0])):
x=[]
for j in range(len(a)):
x.append(a[j][i])
b.append(x)
return b
print covert_array(a)
