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

// BEGIN CUT HERE
/*
 * Author:  plum rain
 * score :
 */
/*

 */
// END CUT HERE
#line 11 "ApproximateDivision.cpp"
#include <sstream>
#include <stdexcept>
#include <functional>
#include <iomanip>
#include <numeric>
#include <fstream>
#include <cctype>
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <set>
#include <queue>
#include <bitset>
#include <list>
#include <string>
#include <utility>
#include <map>
#include <ctime>
#include <stack>

using namespace std;

#define clr0(x) memset(x, 0, sizeof(x))
#define clr1(x) memset(x, -1, sizeof(x))
#define pb push_back
#define sz(v) ((int)(v).size())
#define all(t) t.begin(),t.end()
#define zero(x) (((x)>0?(x):-(x))<eps)
#define out(x) cout<<#x<<":"<<(x)<<endl
#define tst(a) cout<<a<<" "
#define tst1(a) cout<<#a<<endl
#define CINBEQUICKER std::ios::sync_with_stdio(false)

typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<double> vd;
typedef pair<int, int> pii;
typedef long long int64;

const double eps = 1e-8;
const double PI = atan(1.0)*4;
const int inf = 2139062143 / 2;

class ApproximateDivision
{
    public:
        double quotient(int a, int b, int terms){
            int t = 1;
            while (t <= b){
                if (t == b) return (double)a / b;
                t *= 2;
            }
            int c = t - b, cnt = 0;
            double t1 = 1.0, t2 = t; 
            double sum = 0;
            while (cnt < terms){
                sum += a * t1 / t2;
                t1 *= c; t2 *= t;
                ++ cnt;
            }
            return sum;
        }
        
// BEGIN CUT HERE
    public:
    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(); }
    private:
    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(); }
    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 << "	Expected: "" << Expected << '"' << endl; cerr << "	Received: "" << Received << '"' << endl; } }
    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)); }
    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)); }
    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)); }
    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)); }
    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)); }
    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)); }
    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)); }

// END CUT HERE

};

// BEGIN CUT HERE
int main()
{
//    freopen( "a.out" , "w" , stdout );    
    ApproximateDivision ___test;
    ___test.run_test(-1);
       return 0;
}
// END CUT HERE