本文主要是介绍HDU5515 Game of Flying Circus(二分),希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
题意:题解有翻译,然后自己拦截对手时候可以任意走,当然是直线最快啦
题解:http://www.cnblogs.com/qscqesze/p/4931912.html
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define X first
#define Y second
#define cl(a,b) memset(a,b,sizeof(a))
typedef pair<int,int> P;
const int maxn=100005;
const LL inf=1<<27;
double f(double x){return x*x;}
int main(){int T,cas=1;scanf("%d",&T);while(T--){double time,v1,v2;scanf("%lf%lf%lf",&time,&v1,&v2);if(v1==v2){printf("Case #%d: Yes\n",cas++);continue;}if(2*v1*v1>v2*v2){//可以在[2,3)之间拦截double l=0,r=300,m;for(int i=0;i<500;i++){m=(l+r)/2;if(sqrt(f(300)+f(m))/v1>(300+m)/v2){l=m;}else {r=m;}}if(sqrt(f(l)+f(300))/v1+l/v1+2*300/v1<=time+900/v2){printf("Case #%d: Yes\n",cas++);}else {printf("Case #%d: No\n",cas++);}}else if(3*v1>v2){//在[3,4)之间拦截double l=0,r=300,m;for(int i=0;i<100;i++){m=(l+r)/2;if(sqrt(f(300)+f(m))/v1>(900-m)/v2){r=m;}else {l=m;}}if(sqrt(f(300)+f(300-l))/v1+sqrt(f(300)+f(l))/v1+900/v1<=time+1200/v2){printf("Case #%d: Yes\n",cas++);}else {printf("Case #%d: No\n",cas++);}}else{printf("Case #%d: No\n",cas++);}}return 0;
}
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