Black Myth Wukong and The History Behind
Games

Black Myth Wukong and The History Behind

Game Official Website (Chinese) Note: All Chinese names in this article are given in form of \<Given name> \<Family name>. For example, in Sun Wukong, Sun is the family name, Wukong is the given name. Black Myth : Wukong (Chinese: 黑神话:悟空, Pinyin: Hēishénhuà : Wùkōng) is a Chinese game by Game Science Interactive Technology Ltd., short for Game Science. The studio released the trailer of the game on Aug. 20th (CST). And get very much attentions. The official trailer video had got 20 million views on Bilibili. Official Trailer The trailer shows something very impressive. If you hasn't watched it yet, please watch it above first. I'm pretty sure that…

About Infinite Values
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About Infinite Values

Single Variable INF = 0x7fffffff The largest value of 32 bit signed int, which is $2^{32} - 1 = 2147483647 \approx{2.147 \times 10^{9}}$ LLINF = 0x7fffffffffffffff (15 fs) The largest value of 64 bit signed int, which is $2^{64} - 1 = 9223372036854775807 \approx{9.223 \times 10^{18}}$ memset() memset() works by bytes. For example, in C++, one int variable takes a room of 4 bytes. So when int var; memset(var, 0xab, sizeof(var)); and var == 0xabababab。 If it is an array, it works by every byte in every element in the array. It is the same as above. Common Settings memset(a, 0x3f, sizeof(a)) Similar to for(int i = 0; i <…

Some Templates
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Some Templates

Directory Data Structure Trie Tree Segment Tree On Graph Tarjan Tarjan Strongly Connected Components Tarjan Shrink Point Tarjan Cut Point and Bridge Shortest Path Floyd Dijkstra Numbers RMQ Problem by Sparse Table Tree Binary Search Tree Splay Diameter of A Tree Tree Chain Patition Data Structure Trie Tree #include<cstdio> #include<cstring> typedef long long llint; const bool DEBUG_ENABLE = 0; const int MAXN = 1e6 + 5; const int MOD = 19260827; struct TRIE { int nxt[MAXN][30], cnt; bool end[MAXN]; void insert(char str[]) { int now = 0, len = strlen(str); for(int i = 0; i < len; ++i) { int c = str[i] - 'a'; if(!nxt[now][c]) nxt[now][c] = ++cnt; now…