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传送门 感觉最近写代码的状态有点迷...还好这次最后两分钟过了D，不然就掉分了QAQ。 A. Heating 签到。 Code /* * Author: heyuhhh * Created Time: 2019/11/27 21:53:49 */ #include <iostream> #include <algorithm> #include <vector> #include <cmath> #include <set> #include <map> #include <iomanip> #define MP make_pair #define fi first #define se second #define sz(x) (int)(x).size() #define all(x) (x).begin(), (x).end() #define INF 0x3f3f3f3f #define Local #ifdef Local #define dbg(args...) do { cout << #args << " -> "; err(args); } while (0) void err() { std::cout << '\n'; } template<typename T, typename...Args> void err(T a, Args...args) {

直接插入排序 目录 简述 步骤 代码 back 简述 直接插入排序是一种简单的排序，基本操作就是，从一堆数中选择一个数，然后插入到排好序的列表中（本列表）。这种方法也称归位。直接插入排序，每一次选择数字，都是一种归位。 举个例子，就好比如，你在打扑克，然后你抽到了一堆扑克牌，你要对它进行排序，你会很直接地，从一边或者中间选择一张扑克牌，然后根据数值大小，进行插入排序。选出的牌，你会很自然的插入到比它小的前面。不一会，你的手中的扑克牌就完成了排序。 算法的步骤 以下是《数据结构C语言版》的算法步骤： 设待排序的记录存放在数组r[1..n]中，r[1]是一个有序序列。 循环n-1c次，每次使用顺序查找法，查找r[i] (i=2,...n)在已排好序的序列r[1...i-1]中的插入位置，然后将r[i]插入表长为i-1的有序序列r[1...i-1]，直到r[n]插入表长为n-1的有序序列r[1...n-1],最后得到一个表长为n的有序序列。 我的理解： 按要求选出一个最大或者最小的值，然后放到最前面，再从剩下的几个选出最大或者最小的，重复刚刚的操作，一直到整个数组有序 特性 直接插入法的平均时间复杂度是O（n^2^）,其空间复杂度是S（1）。 主要特点是： 是稳定的 算法比较简便，而且容易书写 适用于链表的结构 对基本有序的情况的话，更为适合使用。但是，如果基数非常大，那么复杂度比较高

I have an array of double pointers, but every time I try do print one of the values the address gets printed. How do I print the actual value? cout << arr[i] ? cout << &arr[i] ? they both print the address Does anyone know? If it's really an array of (initialized) double pointers, i.e.: double *arr[] = ... // Initialize individual values all you need is: cout << *arr[i]; cout << *(arr[i]) will print the value. cout << *(arr[i]); If "arr" is declared as double* arr[..]; Then you would use: cout << *(arr[i]) 来源： https://stackoverflow.com/questions/2485565/c-print-value-of-a-pointer

erase()是对string类型的字符串进行删除元素操作的函数 1、erase(int index) 删除下标从index开始直到字符串结尾的元素 1 string s = "1232157"; 2 s.erase(3); 3 cout << s;//123 2、erase(int index,int num) 删除下标从index开始的num个元素 string s = "1234567"; s.erase(3,2); cout << s;//12367 3、 erase(string::iterator it) 删除迭代器指向的元素，函数的返回值是指向删除元素的下一个元素的迭代器 string::iterator it; string s = "1234567"; it=s.erase(s.begin()+1); cout << s << endl;//134567 cout << *it;//3 4、erase(string::iterator it1,string::iterator it2) 删除[it1,it2)区域的元素，函数的返回值是指向删除元素的下一个元素的迭代器 string::iterator it; string s = "1234567"; it=s.erase(s.begin()+1,s.end()-1); cout << s << endl;//17

I decided to try Visual Studio 2012 Express today. First thing to do was to write "Hello world!" application, however, I couldn't make it work. I created a Windows console application project, wrote standard code and it resulted in a run-time error. Here's my code: #include <iostream> using namespace std; int main() { cout << "Hello world!" << endl; system("pause"); return 0; } Looks like something is broken (maybe I missed something?). It gets a runtime error at line 7: http://img443.imageshack.us/img443/7497/coutbroken.png Any help please? :) Your CPU is in urgent need of being junked. Your

I am attempting to have the cout buffer flush to view a string before I manipulate it. Ive attempted to flush the buffer with calls to both std::flush() and std::cout.flush() but neither actually flush my output. Only a call to std::endl has successfully flushed the buffer for me. Here is my code std::istringstream stm (game.date()); int day, month, year; char delim = '/'; std::cout << "Date before: " << game.date() << std::flush; // first flush attempt std::cout.flush(); // second flush attempt doesnt work //std::cout << std::endl; // if this is executed the buffer will flush // Split date

HDU 1880 #include <bits/stdc++.h> #define DBG(x) cerr << #x << " = " << x << endl using namespace std; typedef long long LL; const unsigned int KEY = 137; const int BUCKET_NUM = 120000; struct Hash { vector<pair<string, string>> h[BUCKET_NUM]; unsigned int get_hash(const string &s) { unsigned int hs = 0; for (auto c : s) hs = (hs * KEY + c) % BUCKET_NUM; return hs; } void insert(const string &k, const string &v) { unsigned int hs = get_hash(k); h[hs].push_back(make_pair(k, v)); } string *get(const string &key) { unsigned int hs = get_hash(key); for (auto &kv : h[hs]) { if (kv.first == key)