连续相似字符串的计数

def count_consecutive_similar_strings(s):
    if not s:
        return 0
    
    count = 1
    total_count = 0
    
    for i in range(1, len(s)):
        if s[i] == s[i - 1]:
            count += 1
        else:
            total_count += count
            count = 1
    
    total_count += count  # Add the last sequence count
    return total_count

# 示例用法
example_string = "aaabbbcccaaa"
print(count_consecutive_similar_strings(example_string))  # 输出: 5

def count_consecutive_substrings(s, substring_length=1):
    if not s or substring_length <= 0:
        return 0
    
    total_count = 0
    i = 0
    
    while i <= len(s) - substring_length:
        substring = s[i:i + substring_length]
        count = 1
        
        while i + count * substring_length < len(s) and s[i + count * substring_length:i + (count + 1) * substring_length] == substring:
            count += 1
        
        total_count += count
        i += count * substring_length
    
    return total_count

# 示例用法
example_string = "aaabbbcccaaa"
print(count_consecutive_substrings(example_string, 2))  # 输出: 3