我需要將一個數字範圍劃分爲一些長度相同的段。但我無法決定哪種方式更準確。例如: double r1 = 100.0, r2 = 1000.0, r = r2 - r1;
int n = 30;
double[] position = new double[n];
for (int i = 0; i < n; i++)
{
position[i] = r1 + (double)i/
f(x) = (exp(x)-1)/x;
g(x) = (exp(x)-1)/log(exp(x))
從分析角度而言,f(x) = g(x)適用於所有的x。 當x接近0,無論f(x)和g(x)方法1. % Compute y against x
for k = 1:15
x(k) = 10^(-k);
f(k) =(exp(x(k))-1)/x(k);
De(
如果我編譯與命令g++ -std=c++11 Threads.cpp -lpthread -I../Libs/nr30,其中NR30是由http://www.nr.com/提供的庫將以下代碼,我得到沒有錯誤: #include <chrono>
#include <thread>
int main(void) {
/* ... Sadly, No further code in h