We Can

思路：我们只需要算出每个cell在K次turn之后，状态是ON的概率就行了。在一次turn中一个cell被翻转的概率是多少呢？假设这个cell的坐标是(x,y,z)，那么被翻转的概率就是 (x + 1) * (N - x) * (y + 1) * (M - y) * (z + 1) * (P - 1)/ (N * N * M * M * P * P) ， 然后K次中只要是有奇数次翻转，最后就是ON

#include #include #include #include #include #include #include #include #include #include #include using namespace std;#define rep(i,a,b) for(int i=(a);i<(int)(b);++i)#define rrep(i

Problem BWinning StreakInput: Standard InputOutput: Standard Output "You can run on for a long time,sooner or later God'll cut you down."– Traditional folk songMikael likes to gamble,

U  — PolygonTime Limit: 1 secMemory Limit: 32 MBJohn has been given a segment of lenght N, however he needs a polygon. In order to create a polygon he has cut given segment K times at rand