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SRM 440(1-250pt, 1-500pt)
阅读量:5952 次
发布时间:2019-06-19

本文共 3663 字,大约阅读时间需要 12 分钟。

DIV1 250pt

题意:小球从一段折线斜坡上滚下来,告诉所用时间,求重力加速度。

解法:二分答案模拟即可。

tag:二分,simulation

1 // BEGIN CUT HERE  2 /*  3  * Author:  plum rain  4  * score :  5  */  6 /*  7   8  */  9 // END CUT HERE 10 #line 11 "IncredibleMachine.cpp" 11 #include 
12 #include
13 #include
14 #include
15 #include
16 #include
17 #include
18 #include
19 #include
20 #include
21 #include
22 #include
23 #include
24 #include
25 #include
26 #include
27 #include
28 #include
29 #include
30 #include
31 #include
32 #include
33 #include
34 35 using namespace std; 36 37 #define clr0(x) memset(x, 0, sizeof(x)) 38 #define clr1(x) memset(x, -1, sizeof(x)) 39 #define pb push_back 40 #define sz(v) ((int)(v).size()) 41 #define all(t) t.begin(),t.end() 42 #define zero(x) (((x)>0?(x):-(x))
vi; 49 typedef vector
vs; 50 typedef vector
vd; 51 typedef pair
pii; 52 typedef long long int64; 53 54 const double eps = 1e-8; 55 const double PI = atan(1.0)*4; 56 const int inf = 2139062143 / 2; 57 58 class IncredibleMachine 59 { 60 public: 61 vi x, y; 62 double gao(double g) 63 { 64 int n = sz(x); 65 double v = 0, sum = 0; 66 for (int i = 0; i < n-1; ++ i){ 67 double d = sqrt((y[i]-y[i+1])*(y[i]-y[i+1]) + (x[i]-x[i+1])*(x[i]-x[i+1])); 68 double a = g * (y[i] - y[i+1]) / d; 69 double t = (-v + sqrt(v*v+2*a*d)) / a; 70 sum += t; v += a*t; 71 } 72 return sum; 73 } 74 75 double gravitationalAcceleration(vector
X, vector
Y, int T){ 76 x = X; y = Y; 77 int cnt = 0; 78 double l = 0, r = inf; 79 while (cnt <= 1000){ 80 double mid = (l + r) / 2; 81 if (gao(mid) < T) r = mid; 82 else l = mid; 83 ++ cnt; 84 } 85 return l; 86 } 87 88 // BEGIN CUT HERE 89 public: 90 void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); } 91 private: 92 template
string print_array(const vector
&V) { ostringstream os; os << "{ "; for (typename vector
::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); } 93 void verify_case(int Case, const double &Expected, const double &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } } 94 void test_case_0() { int Arr0[] = { 0,6}; vector
Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = { 100,22}; vector
Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 4; double Arg3 = 9.807692307692307; verify_case(0, Arg3, gravitationalAcceleration(Arg0, Arg1, Arg2)); } 95 void test_case_1() { int Arr0[] = { 0,26,100}; vector
Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = { 50,26,24}; vector
Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 4; double Arg3 = 26.743031720603582; verify_case(1, Arg3, gravitationalAcceleration(Arg0, Arg1, Arg2)); } 96 void test_case_2() { int Arr0[] = { 0,7,8}; vector
Arg0(Arr0, Arr0 + (sizeof(Arr0) / sizeof(Arr0[0]))); int Arr1[] = { 10,6,0}; vector
Arg1(Arr1, Arr1 + (sizeof(Arr1) / sizeof(Arr1[0]))); int Arg2 = 7; double Arg3 = 1.1076837407708007; verify_case(2, Arg3, gravitationalAcceleration(Arg0, Arg1, Arg2)); } 97 98 // END CUT HERE 99 100 };101 102 // BEGIN CUT HERE103 int main()104 {105 // freopen( "a.out" , "w" , stdout ); 106 IncredibleMachine ___test;107 ___test.run_test(-1);108 return 0;109 }110 // END CUT HERE
View Code

 

DIV1 500pt

裸的高斯消元求解概率dp。。。。。。

转载于:https://www.cnblogs.com/plumrain/p/srm_440.html

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