- Python
0603阅读程序写输出结果题目
- 2025-6-3 20:36:12 @
10 道阅读程序写输出结果题目
题目 1
# 遍历列表并累加
nums = [1, 3, 5, 7]
sum_result = 0
for num in nums:
sum_result += num
print(sum_result)
答案:__________
题目 2
# for 循环结合 range 步长
for i in range(2, 10, 3):
print(i, end=" ")
答案:__________
题目 3
# while 循环计算累加(1-5)
s = 0
i = 1
while i <= 5:
s += i
i += 1
print(s)
答案:__________
题目 4
# for 循环遍历字符串,遇指定字符 break
word = "Python"
for char in word:
if char == 't':
break
print(char, end="")
答案:__________
题目 5
# while 循环控制输出次数
count = 3
while count > 0:
print("循环执行中", end=" | ")
count -= 1
答案:__________
题目 6
# for 循环嵌套(简化版)
for i in range(1, 3):
for j in range(1, 3):
print(i * j, end=" ")
答案:__________
题目 7
# while 循环结合条件判断(求 1-10 奇数和)
s = 0
i = 1
while i <= 10:
if i % 2 == 1:
s += i
i += 1
print(s)
答案:__________
题目 8
# for 循环遍历列表,跳过部分元素(用 break 模拟)
fruits = ["apple", "banana", "orange", "grape"]
for fruit in fruits:
if fruit == "orange":
break
print(fruit)
答案:__________
题目 9
# while 循环实现倒计时(从 5 到 1)
n = 5
while n > 0:
print(n, end=" ")
n -= 1
答案:__________
题目 10
# for 循环计算 1-10 偶数的乘积(初始值注意)
product = 1
for i in range(2, 11, 2):
product *= i
print(product)
答案:__________
这些题目覆盖循环结构基础语法、流程控制(break
)、累加 / 累乘逻辑,做完后可对照知识点复盘执行流程,强化对 Python 循环的理解~
1 条评论
-
admin SU @ 2025-6-3 20:36:43
✏️ 阅读程序写输出结果(请写出每段代码的运行结果)
题目1:
x = 5 if x > 3: print("Hello") print("World")
输出结果:
题目2:
num = 7 if num % 2 == 0: print("Even") else: print("Odd")
输出结果:
题目3:
score = 85 if score >= 90: print("A") elif score >= 80: print("B") elif score >= 70: print("C") else: print("D")
输出结果:
题目4(新增循环题):
for i in range(3): for j in range(2): print(i, j)
输出结果:
题目5:
fruits = ["apple", "banana", "cherry"] for fruit in fruits: print(fruit) if fruit == "banana": break else: print("循环结束")
输出结果:
题目6:
total = 0 for i in range(1, 6): total += i print(total)
输出结果:
题目7:
count = 1 while count <= 5: print(count) count += 1
输出结果:
题目8:
i = 1 while True: print(i) i += 1 if i > 3: break
输出结果:
题目9:
for i in range(1, 6): if i % 2 == 0: print(i, end=" ")
输出结果:
题目10:
a = 5 price = 0 if a <= 3: price = 11.0 else: price = 11.0 + (a - 3) * 2.4 print("费用:", price)
输出结果:
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