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I realize this error is usually due to some syntax or type issues but I am not sure how to solve this problem. I think it may do with the type of findRt. vector<triangle> findRightTriangles(unsigned long l, unsigned long h) { <triangle> retval; // storage for return value. triangle t; double what; for(t.s1 = 3; t.s1 <= h; t.s1++) { for(t.s2 = t.s1; t.s2 <= h; t.s2++) { what = sqrt((t.s1*t.s1) + (t.s2*t.s2)); t.s3 = static_cast<unsigned int>(what); if(((t.s1*t.s1)+(t.s2*t.s2)) != (t.s3*t.s3) || t.s1+t.s2+t.s3 < l || t.s1+t.s2+t.s3 > h) { continue; } else if(t.s1+t.s2+t.s3 <= h && t.s1+t.s2+t.s3

Putting together these questions: How can I pad an int with leading zeros when using cout << operator? Printing the correct number of decimal points with cout How can I stream to std::cout, for example, this variable double x = 7.1224 And make it look like this? 07.12 Combine std::setw , std::setfill , std::fixed and std::setprecision : std::cout << std::setfill('0') << std::setw(5) << std::fixed << std::setprecision(2) << x; So, the value for setw is: 2 for the precision desired + 2 for the integer desired + 1 for the floating point. Note: x = 107.1224 will be outputted as 107.12 . 来源： https:

题目链接 简要题意 ： 删除若干个点（不能是所有或空集），满足剩下的点度数在 \(\pmod 3\) 意义下不变。保证图联通。 按照 \(\mod 3 = 0, 1,2\) 分为三组点 \(Z,A,B\) 首先把一些简单的情况判断掉： 有 \(Z\) 点。保留这个点即可。 有两个 \(A\) 点之间右边。保留这两个点即可。 现在只有两类点， \(A\) 类内部不能连。 考虑 \(B\) 类内部连边，如果有环把环提出来即可。（如果整个图是一个大环则无解） 于是 \(B\) 类只能是一棵树。 将 \(B\) 分为 \(\text{A点的个数}\) 类，分别为 \(B_1,B_2,B_3,\dots\) 。（一个点可能属于多个类） 如果 \(B\) 有多种类型，一定有一条路径满足头尾分别在不同的类里面，并且中间的都不在这两类点之中。 于是剩下的点一定是一个森林。 于是 \(A\) 类点只能有一个。 仔细分析还可以发现这个点的度数至少是 \(4\) 。（森林至少有 \(2\) 个叶子，这两个叶子一定会连向 \(A\) ，又需要 $ \mod 3 = 1$） 考虑构造一个 \(A\) 点度数是 \(4\) 的情况。 选定一个点作为根并且 \(A\) 连向它，然后往外面连一个点。 这个时候可以拓展出所有树的情况。 考虑称一颗子树的 期望度数 是叶子节点的个数加上已经与之连上的节点个数。

I've recently started learning C++ at university and decided to advance a bit at home. I had the idea of making a program that, given a piece of text, would print out such text character by character with a small delay in-between (as seen in this video SUPER.HOT chat ). I tried to recreate it using a simple procedure: void typer(string text){ for (int i = 0; i < text.length(); i++){ cout << text[i]; usleep(100000); } But when usleep() is set under 103900, it'll start printing out two characters at a time. My intention is to print only 1 at a time but very quickly. Any suggestions? :D You need

1：运算符重载 1 RMB.h 2 #ifndef RMB_H 3 #define RMB_H 4 5 #include <bits/stdc++.h> 6 using namespace std; 7 class RMB 8 { 9 public: 10 RMB(); 11 RMB(int, int); 12 virtual ~RMB(); 13 void display() 14 { 15 cout << "RMB:" << yuan << "."; 16 if(jf < 10) 17 { 18 cout << "0"; 19 } 20 cout << jf << endl; 21 } 22 friend RMB operator+(const RMB&, const RMB&);//作为友元函数的元素符重载 23 friend bool operator>(const RMB&, const RMB&); 24 int Getjf() { return jf; } 25 void Setjf(int val) { jf = val; } 26 int Getyuan() { return yuan; } 27 void Setyuan(int val) { yuan = val; } 28 RMB operator+(const RMB& s)//作为成员函数的运算符重载

I have this simple code #include <QtCore/qdebug.h> #include <QtCore/qcoreapplication.h> #include <iostream> using namespace std; int main(int argc, char **argv) { cout << "pluto" << endl; QCoreApplication app(argc, argv); qDebug() << "pippo" << endl; return app.exec(); //return 0; } I compiled it with MinGw in Eclipse with no errors, but when I run the code no string message appear on the consolle. What is wrong? Thanks. Luca For cout to work on Windows, you need to have CONFIG+=console in the .pro file. It shouldn't have any effect on any other platform, so you can just add it there. You can