易错诊所：五级最容易踩的五个坑

# 易错诊所：错例 → 诊断 → 正确写法

# 病例 1：质数判定漏掉 n < 2 的“门卫”，1 被误判成质数
def is_prime_wrong(n):
    for i in range(2, n):
        if n % i == 0:
            return False
    return True

def is_prime_right(n):
    if n < 2:
        return False
    i = 2
    while i * i <= n:
        if n % i == 0:
            return False
        i = i + 1
    return True

print("病例1：错例判 1 是质数？", is_prime_wrong(1), "｜正确写法：", is_prime_right(1))

# 病例 2：素数筛忘记划掉 0 和 1
def sieve_wrong(n):
    is_p = [True] * (n + 1)
    for i in range(2, n + 1):
        if is_p[i]:
            for j in range(i * i, n + 1, i):
                is_p[j] = False
    return [x for x in range(n + 1) if is_p[x]]

def sieve_right(n):
    is_p = [True] * (n + 1)
    is_p[0] = is_p[1] = False
    for i in range(2, n + 1):
        if is_p[i]:
            for j in range(i * i, n + 1, i):
                is_p[j] = False
    return [x for x in range(n + 1) if is_p[x]]

print("病例2：错例筛到 10：", sieve_wrong(10))
print("      正确筛到 10：", sieve_right(10))

# 病例 3：辗转相除的循环条件写绕了——最清爽就是 while b != 0
def gcd(a, b):
    while b != 0:
        a, b = b, a % b
    return a

print("病例3：gcd(1071, 462) =", gcd(1071, 462), "（条件写成 b != 0 最稳）")

# 病例 4：二分查找写成 left = mid 会死循环，必须 left = mid + 1
def binary_search(nums, target):
    left, right = 0, len(nums) - 1
    while left <= right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        elif nums[mid] < target:
            left = mid + 1      # 错例 left = mid：mid 已查过还留着，死循环
        else:
            right = mid - 1
    return -1

nums = [3, 8, 15, 21, 34]
print("病例4：在", nums, "里找 21，下标 =", binary_search(nums, 21))

# 病例 5：约数求和漏掉搭档 n // i
def proper_sum(n):
    total = 1
    i = 2
    while i * i <= n:
        if n % i == 0:
            total = total + i
            if i * i != n:      # 完全平方数只加一次
                total = total + n // i
        i = i + 1
    return total

print("病例5：220 的真约数和 =", proper_sum(220), "（漏加搭档 n//i 只会算出 33）")