This recurring algorithm needs to be rewritten. In this case, O(n) is complex and memory will have to be constant (about the number increases).Here's an example of how to count and take it out.#include <iostream> #include <gmp.h> #include <gmpxx.h> #include <chrono> using namespace std; mpz_class pp(int n) { if (n < 1 || n >= 10000000) return 0; if (n < 3) return 1; mpz_class a = 1; mpz_class b = 1; mpz_class c = 1; for (int i = 3; i < n; i++) { mpz_class t = a + b + c; c = b; b = a; a = t; } return a; } int main() { auto t1 = std::chrono::steady_clock::now(); cout &lt;&lt; pp(1000000) &lt;&lt; endl; auto t2 = std::chrono::steady_clock::now(); auto d_milli = std::chrono::duration_cast&lt;std::chrono::milliseconds&gt;( t2 - t1 ).count(); cout &lt;&lt; d_milli &lt;&lt; endl; return 0; } The question is, how fast is that? For a millionth element in my car, it took 22 seconds. For 100,000 elements, about 145ms. For 10,000 units, it's 2-4 ms.And after. How do you compromise?g++ -std=c++11 ff.cpp -lgmpxx -lgmp P.S. must not forget to place the library gmp.updI decided to rewrite with a handful of large numbers. The stowage code is not the best, but gmp only loses six to eight times. If you rewrite sse, I think you can pull more. But even for such a simple exercise, I don't think it's bad.#include <iostream> #include <vector> #include <string> #include <iomanip> #include <gmp.h> #include <gmpxx.h> #include <chrono> #define CLEN 8 #define BASE 100000000 class bi { public: bi() { } bi(std::string data) { for (int i = 0; i < data.length(); i+= CLEN) { int r = std::stoi(data.substr(i, CLEN)); m_digits.push_back(r); } } bi(const bi& data) { m_digits = data.m_digits; } bi(int data) { while (data &gt; 0) { m_digits.push_back(data % BASE); data = data / BASE; } } bi&amp; operator=(bi data) { if (this != &amp;data) { m_digits = data.m_digits; } return *this; } friend std::ostream&amp; operator&lt;&lt; (std::ostream&amp; stream, const bi&amp; b); friend bi operator+( const bi&amp; lhs, const bi&amp; rhs ); private: std::vector&lt;int&gt; m_digits; }; std::ostream& operator<< (std::ostream& stream, const bi& b) { for (int i = b.m_digits.size() - 1; i >= 0; i--) { if (i != b.m_digits.size() - 1) { stream << std::setfill('0') << std::setw(CLEN); } stream << b.m_digits[i]; } return stream; } bi operator+( const bi& lhs, const bi& rhs ) { bi result; size_t ls = lhs.m_digits.size(); size_t rs = rhs.m_digits.size(); size_t lm = std::min(ls, rs); result.m_digits.reserve(std::max(ls, rs) + 1); int p = 0; for (size_t i = 0; i &lt; lm; i++) { int c = lhs.m_digits[i] + rhs.m_digits[i] + p; result.m_digits.push_back(c % BASE); p = c / BASE; } if (ls &gt; rs) { for (size_t i = lm; i &lt; ls; i++) { int c = lhs.m_digits[i] + p; result.m_digits.push_back(c % BASE); p = c / BASE; } } if (rs &gt; ls) { for (size_t i = lm; i &lt; rs; i++) { int c = rhs.m_digits[i] + p; result.m_digits.push_back(c % BASE); p = c / BASE; } } if (p &gt; 0) { result.m_digits.push_back(p); } return result; } template <typename T> T pp(int n) { if (n < 1 || n >= 10000000) return 0; if (n < 3) return 1; T a = 1; T b = 1; T c = 1; for (int i = 3; i < n; i++) { T t = a + b + c; c = b; b = a; a = t; } return a; } template<typename T> int test(int n) { auto t1 = std::chrono::steady_clock::now(); std::cout &lt;&lt; pp&lt;T&gt;(n) &lt;&lt; std::endl; auto t2 = std::chrono::steady_clock::now(); auto d_milli = std::chrono::duration_cast&lt;std::chrono::milliseconds&gt;( t2 - t1 ).count(); std::cout &lt;&lt; d_milli &lt;&lt; std::endl; return d_milli; } int main() { int n = 10000; int gmp = test<mpz_class>(n); int my = test<bi>(n); std::cout << "gmp = " << gmp << " my = " << my << std::endl; std::cout << "my/gmp = " << my/gmp << std::endl; return 0; }

At least. ♪ ♪#include <iostream> using namespace std; int main () { int x,y; for(x = 0; x < 200; x++){ for(y = 0; y < 200; y++){ if (x == y) continue; double firstResult = 5.0 / (x + y); double secondResult = 2.0 / (x - y); if (abs(firstResult - secondResult) < 1e-10) { cout << "x= " << x << " y= " << y << endl; return; } } } cout << "fail" << endl; return 0; } Although, it's understandable that with positive results, x and y It doesn't make sense. xsmaller y♪P.S. I hope you didn't solve the equation, but you trained in programming? :

For small matrices, any mathematical package will be managed with an unrestricted arithmetics (protection of all signs to be obtained in the transformation process). It'll be slow, but he'll answer your questions.For large matrices, it's a completely pointless idea, because the memory and time of the bill will grow exponentially.We'll wonder why computer mathematics have been on computers for 60 years and libraries haven't.First, as I've already noticed, it's almost impossible.Secondly, it is not necessary for anyone to do so, as in practice, the recycled system is well known -- in most cases (diffures, interpolation, minimization), it is a square matrix that is more or less legitimized.Hence, the question of improving the dependence of the matrix, first, and the development of the iterative methods of decision-making, second, should be raised.In order to increase conditionality, there are a number of algorithms that, through trivial changes (only reversing lines and columns), increase conditionality by predominating the main diagonal. Realizing these algorithms is in ViennaCL.To date, auction methods, as the most effective, should be identified among the iterative methods. These are methods that build the mooring space - the method of associated gradients, GMRES and their modifications. Their implementation can also be found in ViennaCL.

“Parametrical equation of sphere with centre at point” https://ru.wikipedia.org/wiki/%D0%A1%D1%84%D0%B5%D1%80%D0%B0 Put your limits on the part where the area was cut off. Get the coordinates of camera A0 from two parameters. Make a vector A1 of A0 to point 0.0.0. Make a vector. A2 (x(A0),y(A0),z(A0)+1)♪ A3=Crossing product A1 on A2 will give the enclosure axle. A4=Crossing product A3 on A1 will give another axle. Get three axles for the A1 A3 A4 vector cell, providing orientation to the camera and camera centre A0. Need to normalize.

First, this cycle makes no sense.for (x=n-1;x < 0;x = x-1){ ^^^^^ I understand the meaning of the original. x is positive, and therefore the cycle will never be fulfilled due to conditions x < 0♪There is also no point in these proposals. x = 0; while(a[x] &lt;= n) { c = a[x]*c; x=x+1; } n = n - 1; int d = 1; ^^^^^^^^^ d = 1/c; ^^^^^^^^^ } You're declaring a local variable. d at the end of the cycle body, and this variable is immediately destroyed.Note that to store the calculations in your formula it is necessary to declare the variables as the number with floating comma and to calculate the expressions accordingly. Also consider that whole numbers are limited, and therefore they can only store the factory for limited importance.You have a lot of senseless code in your program, like this one. n = n + 1; ^^^^^^^^^^ int g = n; for (x=n-1;x < 0;x = x-1) { = g; g = g - 1; } n = n - 1; ^^^^^^^^^^ Instead of two offers you could just write int g = n + 1; EDIT: I'm not strong in Java.import java.util.; import java.lang.; import java.io.*; class Demo { private static double calculate( int n ) { double result = 0.0; double factorial = 1.0; for ( int i = 0; i &lt;= n; i++, factorial /= i ) { result += factorial; } return result; } public static void main (String[] args) throws java.lang.Exception { int n; Scanner scan = new Scanner( System.in ); do { n = 0; System.out.print( "Enter a non-negative number: " ); n = scan.nextInt(); } while ( ! ( n &gt; 0 ) ); System.out.println( "The result is equal to " + calculate( n ) ); } } If there is a number 10the result will be as followsEnter a non-negative number: 10 The result is equal to 2.7182818011463845

https://ru.wikipedia.org/wiki/%D0%9F%D0%BE%D1%80%D1%8F%D0%B4%D0%BE%D0%BA_%D0%B1%D0%B0%D0%B9%D1%82%D0%BE%D0%B2 It's about the number. My car uses little-endian.

Use the usual Viewport-- http://www.w3schools.com/css/css_rwd_viewport.asp + For the size of the window, use the style as a percentage, but not as a picel static, e.g. style=width:90%. It's a good idea to set these styles right in your window reference code. It'll accelerate the download of the page.

So you should read it. http://www.ozon.ru/context/detail/id/4721432/ And then get to know. http://www.rsdn.ru/summary/5259.xml ♪ For this to begin, it is sufficient + you must have basic knowledge of the basis of algorithmization and programming. At the reading stage, it is possible to contact Wikipedia in parallel for explanations of an algorithm.

Long story, read it yourself. Well, if there's a equation, x = f(x), along with the roots (probably where we work) f(x) The module was less than one, but the next approach is simply as a function of the previous one.In your version, you can be seenx = (1/(4*ln(x))-1)^2 orx = e^(1/(4(1+sqrt(x)))) Check yourself that the condition is met in the second case. I should like to note that there is an option where a convenience condition is simply not met and it has to be wandered by a hindrance of functions, but it is completely outside the answer.And then it's just like a taboo:using namespace std; inline double sq(double x) { return x*x; } const double eps = 0.00000001; int main(int argc, const char * argv[]) { double x0 = 1.5, // Начальное приближение x1 = 2; for(;;) { x1 = exp(1.0/(4.0*(1+sqrt(x0)))); if (fabs(x1 - x0) < eps) break; x0 = x1; } cout << x1 << endl; }

Approach to the forehead is a passage along the range and verification of each element by means http://ru.wikipedia.org/wiki/%D0%A0%D0%B5%D1%88%D0%B5%D1%82%D0%BE_%D0%AD%D1%80%D0%B0%D1%82%D0%BE%D1%81%D1%84%D0%B5%D0%BD%D0%B0 ♪Another approach is the preliminary definition of the maximum element in the mass, the selection of simple numbers up to the maximum, then the gangway and the comparison with the selected simple numbers.

Frankly, it's smarter than a mere number dealer to a few. sqrt(N) I don't see. Well, then it's a combination of these simple businessmen. It's clear that when we find a simple divider, we share a number on him and start over. It is understandable, however, that the number of all combinations (=in the number of denominators) is simply a product of all stages of simple denominators increased by 1. Like, 360 = 2^3 * 3^2 * 5^1So, the number of businessmen (3+1)*(2+1)*(1+1)=24♪If the numbers are small, it's only possible to override all the dealers. sqrt(N)taking into account that for each such divider, different from one, there is an appropriate divider on the other side of the business. sqrt(N)♪Need a code? :Update: Example - http://ideone.com/ZgcMoz

If values are not included in the long or int range, BigInteger or BigDecimal should be used. Your task will be to: private static BigInteger factorial(long number) { BigInteger result = BigInteger.ONE; for (long i = 1; i <= number; i++) result = result.multiply(BigInteger.valueOf(i)); return result; }

Chisla x, y and others must be Total♪ To say what it is. 3.1415 % 2You think? Or y/2 The third line is not exactly what is meant by the algorithm description. ♪ ♪

1 - NPD is the largest common denominator. The algorithm to look for NPO 2 numbers is known, but I'll bring him back again.long long gcd(long long a,long long b){ while (a && b) if (a > b) a%=b; else b%=a; return a+b; } There's an assumption that НОД(0,0)=0that's often convenient.Now, in order to find some NCDs, it is necessary to sequentially take NCDs from them. long long gcd(const vector<long long>& num){ if (num.size() == 0) return 0; if (num.size() == 1) return num[0]; long long res = gcd(num[0],num[1]); for (auto i=2;i<num.size() && res!=1;i++) //&& res!=1 можно не писать res = gcd(res,num[i]); } to find the numbers in which all elements of a multiplicity are divided Same balance.There are different ways to do that. But you have an idea to move between two and three times wrong. Example: 5 and 9 provide the balance of 1 in the division by 4. But NCD (5.9)=1. If there's more than 1 number in the kit (proposed to be all different(d) The number of such distributors will not exceed the minimum of these numbers. But it's been a long time.Let x = a*p + r, y = b*p + r Then x-y = (b-a)*p Which means sharing. p similar to the rest of the couples. Therefore, the answer is that all businessmen from NLM Variations All the numbers (Ex N^2, but all are not necessary, sufficient between neighbors or 1 to everything, total N-1 grand).Proof. Let G = gcd(delt)♪Well, then, x ≡ r (mod G). y = x + (y - x) => y = x + G*q => y ≡ r + G*q (mod G) => y ≡ r (mod G). That was necessary.Reverse: для любых (x,y) x ≡ r (mod Q) y ≡ r (mod Q) G not | Q => x-y ≡ 0 (mod Q) => x-y делится на Q. Отсюда противоречие с тем что G - gcd();Real realise yourself.Complicity - memory O(n) or O(1) Time O(n * log2(M) + sqrt(M) )where M is the maximum value. sqrt(M) - Imagine the answer, and it's easy to find algorithms without it.

Look. Let f(n) - number of steps of algorithm.Complexity O(h(n)) means that algorithm requires a number of steps not more than h(n)multiplied by some constant. I mean, lim f(n)/h(n) with growing n It's over.So they say, f(n) = Θ(h(n))if f(n) = O(h(n)) and at the same time h(n) = O(f(n))♪ It means the function. f(n) and h(n) have the same pattern of growth.Example: if the real complexity of the algorithm f(n) = 2n^2 + 30n + log(n^4)♪ f(n) = O(n^2), c. 2n^2 + 30n + log(n^4) / n^2 = 2 + 30/n + 4 log n/n^2 -> 2 infinitely growing n♪But at the same time f(n) = O(n^100)♪ Indeed, 2n^2 + 30n + log(n^4) / n^100 = 2/n^98 + 30/n^99 + 4 log n/n^100 -> 0 infinitely growing n♪ See? I mean, O(h(n)) There's like a top assessment of algorithmrithm's complexity, as rough as possible.Next: f(n) = Θ(n^2)♪ Indeed, f(n) / h(n) -> 2 (when we've already found out h(n) / f(n) obviously seeks 0.5.On the other hand, f(n) ≠ Θ(n^100)because n^100 / (2n^2 + 30n + log(n^4)) = n^98 * (n^2) / (2n^2 + 30n + log(n^4)) = n^98 * (1 / (2 + 30/n + 4 log n / n))and this limit is different, i.e. the denominator is limited and n^98 He wants infinity.Thus, Θ More exact Assessment of the complexity of algorithm.Note for purist mathematics: yes, the limit is not necessary, strictly speaking, necessary. lim sup♪

Sign on keydown and keyup♪ In the first place, display a flag of traffic in this direction, in the second, remove it. It's the timer itself to check if the button is pressed. https://jsfiddle.net/fwg53xc8/6/ in chromium or FF (from ES6):~function() { var LEFT = 37, UP = 38, RIGHT = 39, DOWN = 40; var dirs = {[LEFT]: 0, [UP]: 0, [RIGHT]: 0, [DOWN]: 0}; var SPEED = 10; var x = 0, y = 0; $(document).keydown(function(e) { dirs[e.keyCode] = 1; }).keyup(function(e) { dirs[e.keyCode] = 0; }); setInterval(function() { var $div = $("div"); x -= dirs[LEFT] * SPEED; x += dirs[RIGHT] * SPEED; y -= dirs[UP] * SPEED; y += dirs[DOWN] * SPEED; $div.css('transform', `translate(${x}px, ${y}px)`); }, 100); }();div { position: absolute; left: 50%; top: 50%; margin: -.5em -.5em 0 0; background: red; width: 1em; height: 1em; border-radius: 50%; transform: translate(0, 0); transition: transform 100ms linear; }<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div></div>requestAnimationFrame https://ru.stackoverflow.com/q/529457/178988 On the contrary, the vector motion shall be carried out on a plane, acceleration or deceleration as desired, and also in space. You're at the most primitive level now.Again. I have a very specific case. There's a cage field with all the objects in the cells. For one move, the player either stays in the current cell or moves into one of the eight neighbors. In this way, I'm not interested in the constant speed of the diagonal movement, I'm interested in moving from cell to cell.

Number 0.3 Count to the larger whole. Try 0.3. Check:1,73 / 0,3 = 5,76 ceil(5,76) = 6 6 * 0.3 = 1.8 There's a complete difference, and if the balance is more than 0, add 0,3 - остаток to the reference number. Check 0 is needed to avoid rounding, for example, 1.5 up to 1.8. Check:1,73 % 0,3 = 0,23 0,23 != 0.0 0,3 - 0,23 = 0,07 1,73 + 0,07 = 1,8

As I see it, or I don't know the purpose, and I think it's all elementary:(1) See the average values of the two equal time intervals (or three intervals, to reduce the probability of situation where the splash is one after the other)(2) Comparison and take less(3) Set a rule of the difference in flashing, an example: if the amplitude is three or more times (there is a need to develop the exceedance value) than the lower average found is a spike. On schedule, how will it work (objectives): I take time intervals 100.200 and 200.300. In the first intermediary, the average value of 0.1 (referring only significant figures, i.e. to the comma), the second interval falls into a spike and the mean value increases to about 0.25. We take the value of 0.1 and compare it to the amplitudes according to the rule. If I can help, I'll be glad.

For exampleatan2(y1 - y0, x1 - x0)

https://ru.wikipedia.org/wiki/%D0%9A%D0%BE%D0%BC%D0%BF%D0%BB%D0%B5%D0%BA%D1%81%D0%BD%D0%BE%D0%B5_%D1%87%D0%B8%D1%81%D0%BB%D0%BE#.D0.90.D0.BB.D0.B3.D0.B5.D0.B1.D1.80.D0.B0.D0.B8.D1.87.D0.B5.D1.81.D0.BA.D0.B0.D1.8F_.D1.84.D0.BE.D1.80.D0.BC.D0.B0 That's what it looks like.(a + ib)(c + id) = (ac - bd) + i(bc + ad) ...где i - мнимая единица, для которой i^2 = -1 ...which is https://ru.wikipedia.org/wiki/%D0%9A%D0%BE%D0%BC%D0%BF%D0%BB%D0%B5%D0%BA%D1%81%D0%BD%D0%BE%D0%B5_%D1%87%D0%B8%D1%81%D0%BB%D0%BE#.D0.93.D0.B5.D0.BE.D0.BC.D0.B5.D1.82.D1.80.D0.B8.D1.87.D0.B5.D1.81.D0.BA.D0.B0.D1.8F_.D0.BC.D0.BE.D0.B4.D0.B5.D0.BB.D1.8C For vectors (in an integrated plane) the single length gives another vector a single length whose argument (the angle with Re+, the horizontal axis right) is the sum of the arguments of the constituents.If they weren't the only ones, the length of the result would be the product of their length.