遇到物理题,千万不要一味的当成物理题去想着推出一个最终结果来,这样ACM竞赛成了物理比赛,出题人就没水平了...往往只需要基础的物理分析,然后还是用算法去解决问题.这题n小于等于200,一看就估计是暴力枚举能过.就枚举角度就行了. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath>…

Crazy Tank Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4562    Accepted Submission(s): 902 Problem Description Crazy Tank was a famous game about ten years ago. Every child liked it. Time f…

题意: 给你N个炮弹的发射速度,以及炮台高度H和L1,R1,L2,R2. 问任选发射角度.最多能有几个炮弹在不打入L2~R2的情况下打入L1~R1 注意:区间有可能重叠. 思路: 物理题,发现单纯的依据V去求X很困难. 这个时候想到暴力枚举角度.for(double i=0; i<=pi; i+=0.0007) 算出能到达的x.然后推断x,统计sum 发现以增长级0.0007弧度 刚刚好能过这道题 反正也是醉了~ 代码: #include"cstdlib" #include&qu…

题意: n个物体从高H处以相同角度抛下,有各自的初速度,下面[L1,R1]是敌方坦克的范围,[L2,R2]是友方坦克,问从某个角度抛出,在没有一个炮弹碰到友方坦克的情况下,最多的碰到敌方坦克的炮弹数. 解法: 枚举角度,将pi/2分成1000份,然后枚举,通过方程 v*sin(theta)*t - 1/2*g*t^2 = -H 解出t,然后 x = v*cos(theta)*t算出水平距离,直接统计即可. 代码: #include <iostream> #include <cstdio&…

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> const double g = 9.8; using namespace std; double h,l1,r1,l2,r2;z double cal(double ang,double v) { double vx,vy; vx = v * sin(ang…

题意不难理解,仔细看题吧,就不说题意了 #include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace std; const double PI=acos(-1.0); const double g=9.8; double V[205]; int main() { //freopen("in.txt","r&quo…

Crazy TankTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5033    Accepted Submission(s): 1017 Problem DescriptionCrazy Tank was a famous game about ten years ago. Every child liked it. Time fli…

abs Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description Given a number x, ask positive integer y≥2, that satisfy the following conditions:1. The absolute value of y - x is minimal2. To prime facto…

Safecracker Problem Description === Op tech briefing, 2002/11/02 06:42 CST ===  "The item is locked in a Klein safe behind a painting in the second-floor library. Klein safes are extremely rare; most of them, along with Klein and his factory, were de…

第一次写计算几何,ac,感动. 不过感觉自己的代码还可以美化一下. 传送门:http://acm.hdu.edu.cn/showproblem.php?pid=5128 题意: 在一个坐标系中,有n个点,从中找到两个互不touch,互不cross的两个矩形(边要和坐标轴平行),使得面积最大. 思路: 枚举每个点,n的四次方,这题有个坑点是“回型矩阵”,是真的坑,然后discuss区里说的十字形矩阵是骗人的. 每次对枚举的四个点进行check,如果两个矩阵的四个点相互不在对方的矩阵中,那么这组就是…