Problem Description Once again, James Bond is fleeing from some evil people who want to see him dead. Fortunately, he has left a bungee rope on a nearby highway bridge which he can use to escape from his enemies. His plan is to attach one end of the…

题目大意:给出k:绳子的劲度系数,l:绳长,s:桥高,w:邦德的质量,g取9.81.绳子弹力=形变量*劲度系数.如果落地速度大于10 则摔死,小于0则飘着空中. 题目思路:根据能量守恒得知:落地的动能=重力势能减少量-绳子弹力做功.如果l>s,则绳子不做功. #include<cstdio> #include<stdio.h> #include<cstdlib> #include<cmath> #include<iostream> #inc…

http://acm.hdu.edu.cn/showproblem.php?pid=1155 Bungee Jumping Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 811    Accepted Submission(s): 349 Problem Description Once again, James Bond is fle…

传送门: Bungee Jumping Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1718    Accepted Submission(s): 729 Problem Description Once again, James Bond is fleeing from some evil people who want to se…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 40030    Accepted Submission(s): 18437 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

胜利大逃亡 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 21759    Accepted Submission(s): 8538 Problem Description Ignatius被魔王抓走了,有一天魔王出差去了,这可是Ignatius逃亡的好机会.魔王住在一个城堡里,城堡是一个A*B*C的立方体,可以被表示成A个B*C的矩…

题目地址:HDU 4968 这题的做法是全部学科的学分情况枚举,然后推断在这样的情况下是否会符合平均分. 直接暴力枚举就可以. 代码例如以下: #include <cstring> #include <cstdio> #include <math.h> #include <algorithm> using namespace std; int main() { int t, n, a, i, tot, j, k, h, i1, j1, k1, h1, i2,…

Problem Description 呃......变形课上Harry碰到了一点小麻烦,因为他并不像Hermione那样能够记住所有的咒语而随意的将一个棒球变成刺猬什么的,但是他发现了变形咒语的一个统一规律:如果咒语是以a开头b结尾的一个单词,那么它的作用就恰好是使A物体变成B物体. Harry已经将他所会的所有咒语都列成了一个表,他想让你帮忙计算一下他是否能完成老师的作业,将一个B(ball)变成一个M(Mouse),你知道,如果他自己不能完成的话,他就只好向Hermione请教,并且被迫听…

题目地址:HDU 4970 先进行预处理.在每一个炮塔的火力范围边界标记一个点. 然后对每一个点的伤害值扫一遍就能算出来. 然后在算出每一个点到终点的总伤害值,并保存下来,也是扫一遍就可以. 最后在询问的时候直接推断就可以,复杂度O(2*n). 代码例如以下: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define maxn 110000 #define…

Pinball Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 568    Accepted Submission(s): 244 Problem Description There is a slope on the 2D plane. The lowest point of the slope is at the origin.…