HDU 5144
题意：在给定的高度和初速度，任意抛使得它水平距离最远。
题解：我们可以很容易的推出 length = v*cos(x)*( sqrt( 2.0*h*g + v*v*sin(x)*sin(x) ) + v*sin(x) ) / g; 然后对 [ 0, π/2 ] 之间的弧度三分查找（凸线图形一般用三分）
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #include <math.h> 5 #include <ctype.h> 6 #include <map> 7 #include <set> 8 #include <string> 9 #include <vector> 10 #include <queue> 11 #include <stack> 12 #include <list> 13 #include <algorithm> 14 #include <iostream> 15 #define PI acos( -1.0 ) 16 using namespace std; 17 typedef pair<int,int> P; 18 typedef long long ll; 19 const double E = 1e-7; 20 21 const double g = 9.8; 22 double h, v; 23 24 double dis( double x ) 25 { 26 double ans = v*cos(x)*( sqrt( 2.0*h*g + v*v*sin(x)*sin(x) ) + v*sin(x) ) / g; 27 return ans; 28 } 29 30 int main() 31 { 32 int T; 33 scanf( "%d", &T ); 34 while( T-- ) 35 { 36 scanf( "%lf%lf", &h, &v ); 37 double r, l; 38 double legth = 0; 39 r = PI / 2.0; 40 l = 0; 41 while( r-l > E ) 42 { 43 double mid = l + ( r-l ) / 3; 44 double midmid = r - ( r-l ) / 3; 45 double ans1 = dis( mid ); 46 double ans2 = dis( midmid ); 47 if( ans1 < ans2 ) {l = mid; legth = max( legth, ans2 ); } 48 else { r = midmid; legth = max( legth, ans1 ); } 49 } 50 printf( "%.2lf\n", legth ); 51 } 52 return 0; 53 }
