题&解
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Orange Boy Can You Solve It Out? Ep.5
思考题 for blue boy Roguelike Tree Statement You are given a tree of N nodes. Each node could be 3 status: passable(.), impassable(X), unknown(?). Now we randomly replace each "unknown" to "passable" or "impassable" then we need to find out the exp…… -
Orange Boy Can You Solve It Out? Ep.4
思考题 for nobody Rougelike Game You are given a graph of N*M grids. Each grid can be a wall 'X',a path'.' or unknown '?' One grid is the start point 'S' and one grid is the end point 'E' Now we change the '?' randomly to 'X' or '.' and run shortes…… -
Red boy can you solve it?
Rong Kong is a filial boy. There are N pears on the table. Both he and his brother can eat them. They both want to eat as much as they can. Each of them weighs a[i]. It costs a[i] time for a boy to eat a pear weighing a[i] and it gives him a[i] sat…… -
[赵宇气物理]经典“气”题
考虑空气阻力的竖直上抛问题 由于这篇blog语言儒雅随和通俗易懂,所以一般通过观众也可以放心观看 阻力 f=mkv^2, 其中m为物体质量,k为常数 阻力始终与运动方向相反 重力加速度为g 以初速度 v_0 将该物体上抛,试描述其运动速度与时间的关系 Answer: 首先因为阻力的方…… -
[小题狂做]欧拉公式与暴力展开(1)
由于这篇blog语言儒雅随和通俗易懂,所以一般通过观众也可以放心观看 例题(著名题): $\int^{\infty}{0}{\frac{{e^{cosx}}{sin(sinx)}}{x}}dx这道题乍一看很复杂,但其实在回顾过3B1B的傅里叶变换后,才发现这是一道可以用傅里叶级数暴力的题 首先介绍此题要用到的…… -
Orange Boy Can You Solve It Out? EP.3
思考题 for everyone Man on Graph Given a graph of N nodes and M edges. On node i there are A_i men. Then each person moves along 1 edge randomly connected to the node he stands. But if two men walked along the same edge, they'll fight and both die.…… -
Orange Boy Can You Solve It Out? Ep.2
Solved! Difficulty: D2E 思考题 for everyone Man on Graph You are given M chains each of N+1 nodes. We call the i-th node on the j-th chain (j,i-1). We define a bridge edge connecting (a,b) and (c,d) as follows: - a≠c - b=d - b and d don't equ…… -
Orange Boy Can You Solve It Out? Ep.1
Solved! Difficulty: Div2E 思考题for everyone Dynamic DP You are given: integer arrays A and dp of length N. integer arrays l and r of length N. For each 2\leq i\leq N,1\leq l_i\leq r_i\leq i-1 Known: dp_1=A_1 dp_i=min(dp[l_i,r_i])+A_i for 2\l…… -
回应赵宇“气”的问题
我佛了,空气阻力太难力 111 空气阻力F=kv^2 所以1° 上升时 F=-kv^2-mg dv/dt=-kv^2/m-g ∫(-1/kv^2/m+g)dv=t+v0 左边应该是arctan,所以v=-tanCt+v0,,, 2° 下降时 F=mg-kv^2 ∫(1/mg-kv^2)dv=t+C 有理分式……