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[英]python program, Write a program to print first x terms of the series 3N + 2 which are not multiples of 4
[英]Write a program to print first x terms of the series 3N + 2 which are not multiples of 4
n = int (input())
for x in range (1, n + 1, 1):
for y in range (1, 100, 1):
z = 3 * y + 2
if z % 4 != 0:
print(z, end=' ')
此代碼可以打印不是 4 的倍數的數字,但是如何在打印不是 4 的倍數的前 'n' 個數字 (z) 后停止?
輸入: 10
預期 output: 5 11 14 17 23 26 29 35 38 41
Actual Output: 5 11 14 17 23 26 29 35 38 41 47 50 53 59 62 65 71 74 77 83 86 89 95 98 101 107 110 113 119 122 125 131 134 137 143 146 149 155 158 161 167 170 173 179 182 185 191 194 197 203 206 209 215 218 221 227 230 233 239 242 245 251 254 257 263 266 269 275 278 281 287 290 293 299
使用發電機保持簡單高效。
使用itertools.islice
和itertools.count
:
from itertools import islice, count
N = 10
l = list(islice((x for i in count(start=1) if (x:=3*i+2)%4), N))
output: [5, 11, 14, 17, 23, 26, 29, 35, 38, 41]
引入一個反變量:
n = int (input())
counter = 0
for x in range (1, n + 1, 1):
for y in range (1, 100, 1):
z = 3 * y + 2
if counter >= 10:
break
if z % 4 != 0:
print(z, end=' ')
counter += 1
limit = 10
counter = 0
n = 0
while counter != limit:
n += 1
statement = (3*n)+2
if (statement % 4 != 0) and (statement > 4):
print(statement)
counter += 1
n= int(input())
count=0
x=1
while(count<n):
y=3*x+2
if(y%4!=0):
print(y, end=' ')
count=count+1
x=x+1
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