摘要:
"传送门" Solution 区间DP,枚举断点,对于一个区间,枚举折叠长度,用hash暴力判断是否能折叠即可 Code cpp include include include include include define Re register define F(i,a,b) for(Re int 阅读全文
摘要:
"传送门" Solution 纯搜索80分,加二分90分,再补一个小剪枝满分qwq 真.小剪枝:如果下一个的需求和当前相同,那么不需要再次从头开始试(看代码就明白了233) Code 阅读全文