t和可到达具有单调性,二分就不多说了.下面说下O(1)的做法,实际上是等效转换,因为答案一定存在,如果在t0之前,那么分解一下 直接按照只有v计算就可以了.反过来如果计算的结果大于t0,那么表示答案在t0之后.因为速度分量是可以独立累加的,因此 可以找到一开始就只有w的等效的点. #include<bits/stdc++.h> using namespace std; ], y[]; double vm, t0; ], w[]; ; double dex, dey, D; bool uniti…

这题可以o(1)推出公式,也可以二分答案+验证. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> using namespace std; double X1,X2,Y1,Y2; double v,t; double vx,vy,wx,wy; bool check(double mid) { double tmpx,…

D. Chip 'n Dale Rescue Rangers Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/problem/D Description A team of furry rescue rangers was sitting idle in their hollow tree when suddenly they received a signal of distress.…

题目链接: 题目 B. Chip 'n Dale Rescue Rangers time limit per test:1 second memory limit per test:256 megabytes 问题描述 A team of furry rescue rangers was sitting idle in their hollow tree when suddenly they received a signal of distress. In a few moments they…

题目链接: B. Chip 'n Dale Rescue Rangers time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A team of furry rescue rangers was sitting idle in their hollow tree when suddenly they received a signa…

B. Chip 'n Dale Rescue Rangers time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output A team of furry rescue rangers was sitting idle in their hollow tree when suddenly they received a signal of d…

题意:从一点到另一点,前t秒的风向与t秒后风向不同,问到另一点的最短时间 挺难的,做不出来,又参考了别人的代码.先得到终点指向起点的向量,设T秒钟能到.如果T>t则受风1作用t秒,风2作用T-t秒,否则只受风1作用t秒.风作用对向量的影响加到向量上.如果该向量长度小于(船速*T)^2的话,说明T秒能到.T实际上用low和high的mid得出,如果T秒可行,则提高下限. 乱码: //#pragma comment(linker,"/STACK:1024000000,1024000000&qu…

A. Wizards' Duel Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/problem/A Description Harry Potter and He-Who-Must-Not-Be-Named engaged in a fight to the death once again. This time they are located at opposite ends of…

CodeForces 591A 题意:在距离为L的两端A,B,相向发射魔法,a(以P1的速度)-->B,A<--b(以P2的速度).假设a-->B,途中相遇,则返回到原点A<--a. 后又继续,a-->B,速度不变. b亦是如此.求第二次相遇时a的位移. 思路:因为速度不变,所以第二次相遇地点与第一次相遇地点一样. res= n/(Va+Vb)*Va 代码: #include <iostream> #include <cstdio> #include…

E. Three States Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/problem/E Description The famous global economic crisis is approaching rapidly, so the states of Berman, Berance and Bertaly formed an alliance and allowed…