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Time Limit: 20 Sec
Memory Limit: 256 MB
http://acm.hdu.edu.cn/showproblem.php?pid=5515
Input
The first line contains an integer t (0<t≤1000), followed by t lines.
Each line contains three double T, V1 and V2 (0≤V1≤V2≤2000,0≤T≤2000) with no more than two decimal places, stands for one case.
Output
If there exist any strategy for Asuka to win the match, output ``Yes", otherwise, output ``No".
Sample Input
Sample Output
Case #1: No
Case #2: Yes
题意
题解：
http://blog.csdn.net/snowy_smile/article/details/49535301
这个文章讲的非常完美！
代码
#include<iostream> #include<stdio.h> #include<math.h> using namespace std; int main() { int t;scanf("%d",&t); for(int cas=1;cas<=t;cas++) { double T,v1,v2; cin>>T>>v1>>v2; if(v1==v2)//第一段相遇 { printf("Case #%d: Yes\n",cas); continue; } if(300.0 * sqrt(2) / v1 < 600.0 / v2)// 第二段相遇 { double l = 0,r = 300; for(int i=0;i<100;i++) { double mid = (l+r)/2.0; double len = mid * mid + 300 * 300; len = sqrt(len); double t1 = len / v1; double t2 = 300 / v2 + mid / v2; if(t1 > t2)l = mid; else r = mid; } double t1,t2; t1 = sqrt(l*l+300*300)/v1+l/v1+2*300/v1; t2 = T + 3*300 / v2; if(t1<=t2)printf("Case #%d: Yes\n",cas); else printf("Case #%d: No\n",cas); } else if(300.0 / v1 < 900.0 / v2) { double l = 0,r = 300; for(int i=0;i<100;i++) { double mid = (l+r)/2.0; double len = mid * mid + 300 * 300; len = sqrt(len); double t1 = len / v1; double t2 = (900 - mid) / v2; if(t1 > t2)r = mid; else l = mid; } double t1,t2; t1 = sqrt(300*300+l*l)/v1+sqrt(300*300+(300-l)*(300-l))/v1+3*300/v1; t2 = T + 300*4/v2; if(t1<=t2)printf("Case #%d: Yes\n",cas); else printf("Case #%d: No\n",cas); } else printf("Case #%d: No\n",cas); } }
HDU 5515 Game of Flying Circus 二分
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原文地址：http://www.cnblogs.com/qscqesze/p/4931912.html
