读程序,总结程序功能
![](https://img.haomeiwen.com/i14187156/94fcfb0284627829.png)
求2的20次方
2.
![](https://img.haomeiwen.com/i14187156/371410494fc1a5cb.png)
计算出1到100中,能被3和7整除,但不是3和7的倍数的数有多少个
编程实现题(用for和while各写一遍):
1.求一到100之间所有数的和、平均值
sum = 0
num = 1
for num in range(1,101):
sum += num
num += 1
print('和:%d 平均值:%.2f'%(sum,sum/100))
# 和:5050 平均值:50.50
sum1 = 0
num = 1
while num <= 100:
sum1 += num
num += 1
print('和:%d 平均值:%.2f'%(sum1,sum1/100))
#和:5050 平均值:50.50
2.计算1-100之间能3整除的数的和
sum1 = 0
num = 1
for num in range(0,101):
if num%3 == 0:
sum1 += num
num += 1
print(sum1)
# 1683
sum1 = 0
num = 1
while num <= 100:
if num % 3 == 0:
sum1 += num
num += 1
print(sum1)
#1683
3.计算1-100之间不能被7整除的数的和
sum1 = 0
num = 1
for num in range(0,101):
if num % 7 == 0:
sum1 += num
num += 1
print(sum1)
#4315
sum1 = 0
num = 1
while num <= 100:
if num % 7 != 0:
sum1 += num
num += 1
print(sum1)
#4315
1.求斐波那契数列中第n个数的值:1,1,2,3,5,8,13,21,34....
n = 4
n_1 = 1 #前一个
n_2 = 1 #前两个
if n <= 2: #n<=2时候用
print(1) #n<=2时候用
current = 1 #n<=2时候用
for x in range (3,n+1):
# 算出后一位
current = n_1 + n_2
# 往后移
n_2 = n_1
n_1 = current
print(current)
# 3
判断101-200之间有多少个素数,并输出所有素数。判断素数的方法:用一个数分别除2到sqrt(这个数),如果能被整除,则表明此数不是素数,反之是素数
for x in range(101,201):
count = 0
# y取2~x-1
for y in range (2,x):
if x % y == 0:
count += 1
print('%d不是素数'% (x))
break
if count == 0:
print('%d是素数' % (x))
- 打印出所有的水仙花数,所谓水仙花数是指一个三位数,其各位数字立方和等于该数本身。 如:153是个水仙花数,因为153 = 1^3 + 5^3 + 3^3
for x in range(100,1000):
ge_wei = x%10
shi_wei = x//10%10
bai_wei = x//100
if x == ge_wei**3 + shi_wei**3 + bai_wei**3:
print('%d是水仙花数'%x)
4.有一分数序 :2/1,3/2,5/3,8/5,13/8,21/13...求出这个数 的第20个分数 分子:上 个分数的分 加分 分 : 上 个分数的分 fz = 2 fm = 1 fz+fm / fz
fz = 2
fm = 1
for x in range(20):
fz,fm = fz+fm,fz # fz,fm = 3,2
print('%d/%d'%(fz,fm))
# 28657/17711
- 给一个正整数,要求:1、求它是几位数 2.逆序打印出各位数字
num = 16723 #整除10
num2 = num
count = 0
while num != 0:
count+=1
num //= 10
print(count)
# 6
num = 16723
str_num = str(num)
print(str_num[::-1])
# 32761
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