下一个更高的数字，具有相同的设置位数和固定的一些位

def find_next_number(original_number, digit_positions, new_values):
    number_str = str(original_number)
    for position, new_value in zip(digit_positions, new_values):
        number = int(number_str)
        power = len(number_str) - position - 1
        number -= (number // (10 ** power)) % 10 * (10 ** power)
        number += new_value * (10 ** power)
        number_str = str(number)
    return int(number_str)

# 示例用法
original_number = 12345
digit_positions = [2, 3]  # 需要修改的位数，这里是第二位和第三位
new_values = [6, 7]  # 对应位数需要修改的值，这里是将第二位修改为6，第三位修改为7
next_number = find_next_number(original_number, digit_positions, new_values)
print(next_number)