SRM 389(1-250pt)
题意:按一定方法生成n个分数,求他们的和。n <= 20
解法:暴力。我只是没想到,10000^20用double算也能被接受0 0
tag:brute-force
1 // BEGIN CUT HERE 2 /* 3 * Author: plum rain 4 * score : 5 */ 6 /* 7 8 */ 9 // END CUT HERE 10 #line 11 "ApproximateDivision.cpp" 11 #include <sstream> 12 #include <stdexcept> 13 #include <functional> 14 #include <iomanip> 15 #include <numeric> 16 #include <fstream> 17 #include <cctype> 18 #include <iostream> 19 #include <cstdio> 20 #include <vector> 21 #include <cstring> 22 #include <cmath> 23 #include <algorithm> 24 #include <cstdlib> 25 #include <set> 26 #include <queue> 27 #include <bitset> 28 #include <list> 29 #include <string> 30 #include <utility> 31 #include <map> 32 #include <ctime> 33 #include <stack> 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))<eps) 43 #define out(x) cout<<#x<<":"<<(x)<<endl 44 #define tst(a) cout<<a<<" " 45 #define tst1(a) cout<<#a<<endl 46 #define CINBEQUICKER std::ios::sync_with_stdio(false) 47 48 typedef vector<int> vi; 49 typedef vector<string> vs; 50 typedef vector<double> vd; 51 typedef pair<int, int> 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 ApproximateDivision 59 { 60 public: 61 double quotient(int a, int b, int terms){ 62 int t = 1; 63 while (t <= b){ 64 if (t == b) return (double)a / b; 65 t *= 2; 66 } 67 int c = t - b, cnt = 0; 68 double t1 = 1.0, t2 = t; 69 double sum = 0; 70 while (cnt < terms){ 71 sum += a * t1 / t2; 72 t1 *= c; t2 *= t; 73 ++ cnt; 74 } 75 return sum; 76 } 77 78 // BEGIN CUT HERE 79 public: 80 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(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); if ((Case == -1) || (Case == 5)) test_case_5(); if ((Case == -1) || (Case == 6)) test_case_6(); } 81 private: 82 template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); } 83 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; } } 84 void test_case_0() { int Arg0 = 2; int Arg1 = 5; int Arg2 = 2; double Arg3 = 0.34375; verify_case(0, Arg3, quotient(Arg0, Arg1, Arg2)); } 85 void test_case_1() { int Arg0 = 7; int Arg1 = 8; int Arg2 = 5; double Arg3 = 0.875; verify_case(1, Arg3, quotient(Arg0, Arg1, Arg2)); } 86 void test_case_2() { int Arg0 = 1; int Arg1 = 3; int Arg2 = 10; double Arg3 = 0.33333301544189453; verify_case(2, Arg3, quotient(Arg0, Arg1, Arg2)); } 87 void test_case_3() { int Arg0 = 1; int Arg1 = 10000; int Arg2 = 2; double Arg3 = 8.481740951538086E-5; verify_case(3, Arg3, quotient(Arg0, Arg1, Arg2)); } 88 void test_case_4() { int Arg0 = 1; int Arg1 = 7; int Arg2 = 20; double Arg3 = 0.14285714285714285; verify_case(4, Arg3, quotient(Arg0, Arg1, Arg2)); } 89 void test_case_5() { int Arg0 = 0; int Arg1 = 4; int Arg2 = 3; double Arg3 = 0.0; verify_case(5, Arg3, quotient(Arg0, Arg1, Arg2)); } 90 void test_case_6() { int Arg0 = 50; int Arg1 = 50; int Arg2 = 1; double Arg3 = 0.78125; verify_case(6, Arg3, quotient(Arg0, Arg1, Arg2)); } 91 92 // END CUT HERE 93 94 }; 95 96 // BEGIN CUT HERE 97 int main() 98 { 99 // freopen( "a.out" , "w" , stdout ); 100 ApproximateDivision ___test; 101 ___test.run_test(-1); 102 return 0; 103 } 104 // END CUT HERE
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现在的你,在干什么呢?
你是不是还记得,你说你想成为岩哥那样的人。
现在的你,在干什么呢?
你是不是还记得,你说你想成为岩哥那样的人。