Solutions
我的方法是枚举n+m的值，对于每个(n+m)%5==0，分类讨论，得出数对数，累加
排序扫一遍，贪心尽量增大mex值
三遍DFS，分别求某节点上的最大C和、子节点数、累加结果
DP，加一维表示是否连续
如果有0，有1、2、3即可 如果没有0，矩形覆盖了0以外的部分即可
建边权为1的图，跑bfs或spfa即可
函数f(N) = N/2^3 + N/3^3 + N/4^3 + N/5^3 + ... 显然有单调性，二分即可
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