按一个按钮两次相当于没按，每个按钮按或者不按的情况，一共有2^4=16种状态。枚举每个按钮是否按下，及时根据按键总数终止不可能的情况，注意按钮1==按钮2+按钮3的作用，因此在c 4==2时注意sum！=1的情况。虽然灯最多有100盏,但6盏以后的灯都是前6盏的重复，因此只需保存前六盏的状态。最后对所有合适的情况排序输出。-Equivalent to not press each button press or by pressing a button twice, a total of 2 ^ 4 = 16 states. Enumerate each button is pressed to terminate the unlikely event that a timely manner according to the total number of keys, pay attention to the role of the button 1 == the button 2+ button, c　4 == 2 Note the sum = 1. Although the lights up to 100, after six lights are the top six replicates, so just save the six states. Finally, all the right sort output.