2012 JinHuaSite Regional Contest Problem D
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此题可以用三分法(见网上相关解法),也可以暴力枚举,这就是为什么很多人可以过这题的原因。
暴力枚举角度即可,至于其中的公式则是简单的物理公式了,算一个抛物线即可。但角度的选取范围要大于 0 - 90 度的范围,个人的理解是,如果坦克在炮的正下方的时候,则炮要把炮头转到x轴下方才可能打到坦克。
难度等级:★★☆☆☆
#include<cstdio>#include<cstring>#include<cstdlib>#include<string>#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<stack>#include<cmath>#include<map>#include<time.h>using namespace std;const int M=1000003;const int N=1001;const int inf=1<<29;//const double inf=1000000000000000000LL;const double eps=1e-12;const double pi=acos(-1.0);int n;double v[N],h,l1,r1,l2,r2;const double g=9.800; int find(double ang){ int tot=0; for(int i=0;i<n;i++) { double vy=v[i]*cos(ang); double t=(vy+sqrt(vy*vy+2*h*g))/g; double l=v[i]*sin(ang)*t; if(l>=l2 && l<=r2) return 0; if(l>=l1 && l<=r1) tot++; } return tot;}int main(){ //printf("%lf\n",cos(1.5707)); while(scanf("%d",&n),n) { scanf("%lf%lf%lf%lf%lf",&h,&l1,&r1,&l2,&r2); for(int i=0;i<n;i++) scanf("%lf",&v[i]); int ans=0; for(double i=0.000000;i<=2.2000;i+=0.0007) { int cnt=find(i); if(cnt>ans) ans=cnt; } printf("%d\n",ans); } return 0;}
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