新曙光

想弄高中信息学竞赛的，过来看看。

// C++ program for the above approach 
#include <bits/stdc++.h> 
using namespace std; 

// MUltiply the number x with the number 
// represented by res array 
vector<int> multiply(long int n, vector<int> digits) 
{ 

	// Initialize carry 
	long int carry = 0; 

	// One by one multiply n with 
	// individual digits of res[] 
	for (long int i = 0; i < digits.size(); i++) { 
		long int result 
		= digits[i] * n + carry; 

		// Store last digit of 'prod' in res[] 
		digits[i] = result % 10; 

		// Put rest in carry 
		carry = result / 10; 
	} 

	// Put carry in res and increase result size 
	while (carry) { 
		digits.push_back(carry % 10); 
		carry = carry / 10; 
	} 

	return digits; 
} 

// Function to recursively calculate the 
// factorial of a large number 
vector<int> factorialRecursiveAlgorithm( 
long int n) 
{ 
	if (n <= 2) { 
		return multiply(n, { 1 }); 
	} 

	return multiply( 
	n, factorialRecursiveAlgorithm(n - 1)); 
} 

// Driver Code 
int main() 
{ 
	long int n = 50; 

	vector<int> result 
	= factorialRecursiveAlgorithm(n); 

	for (long int i = result.size() - 1; i >= 0; i--) { 
		cout << result[i]; 
	} 

	cout << "\n"; 

	return 0; 
}

想弄高中信息学竞赛的，初中就要开始练习！

新曙光188029提示用法：请在编译器中对着行号，看程序分析。

姬里源 我的写法你能看懂不，不懂语音，如果懂，那这步对不对？

紫圈部分？

姬里源：对着呢

新曙光188029：但我还是不明白，这个digits干什么用！

#include <iostream>
#include <vector>
 
int main()
{
    // Create a vector containing integers
    std::vector<int> v = {8, 4, 5, 9};
 
    // Add two more integers to vector
    v.push_back(6);
    v.push_back(9);
 
    // Overwrite element at position 2
    v[2] = -1;
 
    // Print out the vector
    for (int n : v)
        std::cout << n << ' ';
    std::cout << '\n';
}

新曙光188029提示：这是比较复杂的数据类型了。这里面的东西是有序的，不要被大括号弄糊涂了，这不是数学中的集合。

How soon will machines outsmart humans? The biggest brains in AI disagree

// C++ program to check if a 
// number is prime 
#include <iostream> 
#include <math.h> 
using namespace std; 

int main() 
{ 
	int n, i, flag = 1; 

	// Ask user for input 
	cout << "Enter a number: "; 

	// Store input number in a variable 
	cin >> n; 

	// Iterate from 2 to sqrt(n) 
	for (i = 2; i <= sqrt(n); i++) { 

		// If n is divisible by any number between 
		// 2 and n/2, it is not prime 
		if (n % i == 0) { 
			flag = 0; 
			break; 
		} 
	} 

	if (n <= 1) 
		flag = 0; 

	if (flag == 1) { 
		cout << n << " is a prime number"; 
	} 
	else { 
		cout << n << " is not a prime number"; 
	} 

	return 0; 
} 

// This code is contributed by shivanisinghss2110.

新曙光188029提醒同学们注意，循环的上限是根号n，你知道为什么吗？

这段代码可以求复数根：

// C++ program to find roots of
// a quadratic equation
#include <bits/stdc++.h>
using namespace std;

// Prints roots of quadratic equation
// ax*2 + bx + x
void findRoots(int a, int b, int c)
{
    // If a is 0, then equation is
    // not quadratic, but linear
    if (a == 0) {
        cout << "Invalid";
        return;
    }

    int d = b * b - 4 * a * c;
    double sqrt_val = sqrt(abs(d));

    if (d > 0) {
        cout << "Roots are real and different ";
        cout << (double)(-b + sqrt_val) / (2 * a) << " "
             << (double)(-b - sqrt_val) / (2 * a);
    }
    else if (d == 0) {
        cout << "Roots are real and same ";
        cout << -(double)b / (2 * a);
    }

    // d < 0
    else {
        cout << "Roots are complex ";
        cout << -(double)b / (2 * a) << " + i"
             << sqrt_val / (2 * a) << " "
             << -(double)b / (2 * a) << " - i"
             << sqrt_val / (2 * a);
    }
}

// Driver code
int main()
{
    int a = 1, b = -7, c = 12;

    // Function call
    findRoots(a, b, c);
    return 0;
}

// C++ program to find the sum
// of natural numbers up to n
// using recursion
#include <iostream>
using namespace std;

// Returns sum of first
// n natural numbers
int recurSum(int n)
{
	if (n <= 1)
		return n;

	return n + recurSum(n - 1);
}

// Driver code
int main()
{
	int n = 5;
	cout << recurSum(n);

	return 0;
}

不是用for求和，这是理解递归调用的最简单的例子。以下是新曙光188029的小伙伴给出的运行分析：

adobe illustrator, 锚点 anchor，leftDirection and rightDirection

有诗为证：

先运行这个程序:

var docRef = documents.add();
var myPath = app.activeDocument.pathItems.add();

var newPoint = myPath.pathPoints.add();
newPoint.leftDirection = [350, 350];
newPoint.anchor = [200, 200];
newPoint.rightDirection = [350, 200];

var newPoint1 = myPath.pathPoints.add();
newPoint1.leftDirection = [500, 500];
newPoint1.anchor = [500, 500];
newPoint1.rightDirection = [500, 500];

再运行这个程序：

var myPath = app.activeDocument.pathItems.add();
myPath.setEntirePath( new Array(
new Array(500,500),
new Array(350,200)))

我用了10天时间，搞清了左倾和右倾，根本google不到，更别说百度了！

188029部落晚上都不睡觉，独孤电烤查找script说明书的handle，发现了上面一段。说明，前面理解有误，但毕竟与真相接近了一大步。

其实，就是曲线的左切线和右切线，如果你还记得数学分析里讲的左右极限，就是这个味道的！