#include <iostream> #include <chrono> using namespace std; int main() { cout << chrono::high_resolution_clock::period::den << endl; auto start_time = chrono::high_resolution_clock::now(); int temp; for (int i = 0; i< 242000000; i++) temp+=temp; auto end_time = chrono::hi...
#include <iostream>usingnamespace std;// 第一个命名空间namespace first_space{ void func() { cout << "first_space" << endl; }}namespace second_space{ void func() { cout ... c 命名空间之using指令 命名空间 ios #include mysql
For creating DAL Class right click our Helper folder and add new Class file and give the name as SQLDALClass.cs. Copy the below code and paste in the DAL Class.AS we have already seen if the Namespace is different then enter your namespace. Note: In this class file I ...
#include <bits/stdc++.h> using namespace std; #define long long long const int M = 1000000007; map<long, long> F, G; long f(long n) { if (F.count(n)) return F[n]; long n1=n/2, n2=n-n1; return F[n] = (f(n1)*f(n2) + f(n1-1)*3*f(n2-1)) % M; } long g...
血泪史: 博主从前只知道sql优化是:不要用select * ,between and代替<>,in...非常low的优化知识。但自从发现了EXPLAIN之后,才觉得真正进入了sql优化的世界。 正题: 开发中有时会出现一个复杂的sql查询执行时间很长的问题,这个时候就需要通过EXPLAIN来进行分析啦!应用: EXPLAIN的用 ...
using namespace std;//! Buffer that holds block of characters and last character of previous buffer.class MyBuffer {static const size_t buffer_size = 10000;char* my_end;//! storage[0] holds the last character of the previous buffer.char storage[1+buffer_size];public...
using namespace std; const int SIZE = 10; class Bank_Acct { public: Bank_Acct( ); //default constructor Bank_Acct(double new_balance, string Cname); //explicit value //constructor void Print(ostream & out); //accessor function
using namespace std;using namespace tbb; template <class T>class Body { T reduced_result; T* const y; const T* const x; public: Body( T y_[], const T x_[] ) : reduced_result(0), x(x_), y(y_) {} T get_reduced_result() const {return reduced_result;} template<typename ...
using namespace std; bool IsPrime (int); int main(void) { cout<<"The program checks if the given number is prime!"<<endl; for(int i= 0; i<=44; i++, cout<<'_'); cout<<endl; do { cout<<"Do you wish to test next number y/n->"; ...
This program will take a lower limit, higher limit, and dx value as an input and will display the final value of integral as an output. The number of iteration of the for loop will depend upon the input values. Answer and Explanation: #include <iostream> using namespa...