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Battery algorithms

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Oba22 last edited by

What algorithms exist with the use of battlefields initiated in structures?
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inna last edited by
Since this is the case battlefieldsNot battle vectorsa more appropriate example of their use:
```
typedef struct
{
    union
    {
       struct
       {
           unsigned long mantissa: 23;
           unsigned long exponent: 8;
           unsigned long sign: 1;
       } float_parts;
       float all;
    };
} _float __attribute__((__packed__));
```
Here, the use of battlefields allows easy access to the inner content of the number according to IEE754 (Of course there are other problems. endiannessbut this is not so principled)♪
Previous battlefields (e.g. in the first gambling variations idTech) Used frequently enough to save memory of any data.
Now their use, in my view, is pointless, and, in general, access to such a structure takes longer because of the need masking'a♪
More practically, I recommend that the article be read http://www.gotw.ca/publications/N1211.pdf Which explains how bad it is. vector<bool>♪
It's to learn how to use it. std::bitset and boost::dynamic_bitset♪
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Battery operations
Software Programming • c language ponteiro stac languagek • • rielyn

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Most of the error is actually something very simple and even distraction, which is the fact that it is calling the same function more than once:mpy(&p); // <-- mpy aqui printf("\nmultiplicacao: %.1f\n",mpy(&p) /* <-- e aqui*/); add(&p); // <-- add aqui printf("\nadicao: %.1f\n",add(&p) /* <-- e aqui */); And the same happens for all operations that are not push or pop. These functions already return the value soon you just need to keep them in printfs, so:printf("\nmultiplicacao: %.1f\n",mpy(&p)); printf("\nadicao: %.1f\n",add(&p)); To conclude the logic that showed in the example, decreasing was also incomplete because it only returns the decremented element and does not alter it in the stack. In this sense its function dec should be:float dec(struct Pilha *p) { int x; p->pElem[p->topo]--; return p->pElem[p->topo]; } With these two changes I mentioned, you already get the result you expect. https://ideone.com/UHdhab There are several things that can improve in the code, however I leave just a few to consider:The field capa Pilha is not being used, and therefore allows you to enter more elements than the indicated capacity. Even the name itself is not very good, since I only realized so it served when I looked at the function that builds the stack.Similarly, it should not allow to remove elements from the queue with pop if there is none, that is, if topo already -1. The way it is if this situation happens will have indefinite behavior and potential segmentation fault.In operation div must prevent division from being made by 0, which was actually the error I had initially, since the operations were made to double and remove elements more from the stack ending in zero operations.
Complexity of the algorithm of
Software Programming • python time complexity • • carriann

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We will address this issue in general for different languages.Case one. Lines variables, the row is an indication of a set of symbols (or other data structure) (+ possibly folded with functions). Examples of languages - C (char *) and C++ (string). In this case, the code from the question is essentially released into the normal addition of the element to the mass. And if it's not. Low quality laba the difficulty of adding (amortized) will not be more than the length of what is added.Then it is obvious that it will be O(1) and total O(len).The benefits of this approach are the speed of work (addition, possibility of replacement of symbols, fast-tracking). Less - need to be very careful what and how the line changes, especially in multi-point implementation.And two cases, lines. non-replaceable♪ The object of the line shall be created once and shall not alter its condition directly or indirectly. This is typical of Java. In this case, there is every time a temporary facility is being set up to replicate all the old line plus what needs to be added. I mean, rudely.@s - string r[0] = '' for i =1: i <= s.length(); i++ r[i] = String.gen(r[i-1] , s[i-1]); And we create a Len line, long from 0 to Len inclusive, it's hard to see that their total length will be Len*(Len+1)/2 ~ O(Len^2). By the way, in the absence of a proper garbage collector (but usually in such languages), it is very memorable.Pros and minus can be said backwards, working easier but slower.But in every language, there is a mechanism that allows for the creation of such a thing without losing its productivity. This may be a building building of a collection or special utilitar classes (StringBuiled, etc.). Using them has made it possible to speed up the process substantially and to reach the O(len) asymmetry in this example.A smart compililer, if he can prove the equivalence of the code (and in such a simple situation this is not difficult), can replace a direct tilt line content with a speed win. That's what happened in Pitton. http://ideone.com/wqfd8C ♪ But it's dangerous to rely on it. Citates from comments.Try Jython, for example. Here's a detailed description from the language creator. CPython (after 2.5) can sometimes optimize (as a special case) the type r += code. Because it's hard for newcomers to explain that they always have to use ''.join instead of s+=. That's a good answer from Alex Martelli to the question.(Reference to question - https://stackoverflow.com/a/1350289/4279 )To understand the fragility of optimization, compare: python -mtimeit -s 's=a'*10000' $'r =In this case, the compiler lacks "conformity" and he's realising what's written in the "lob," which is much slower. So again, we need to use special language designs and simplify the life of the compiler.
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Realize BFS algorithm
Software Programming • c sharp algorithm • • Saumya

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You have a DFS. LinkedList It's like a glass, you're at the start of the rolling, and you're going around. And if you use it as a turn, you end up instead. AddFirstIt'll be BFS.
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Call recursive function with decreasing operator
Software Programming • c++ operators • • rielyn

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Why use this operator over there? What do you expect to get from this? It will decrease a variable that will not be used later, so it doesn't make sense to do this, it's just making the code inefficient and with potential problem since this is considered https://pt.stackoverflow.com/q/173886/101 . This operator has the side effect of changing the status of the variable, when it occurs at a time that the variable is being used in the same expression is not known the order that will be executed each operation and everything can happen. What you want is just picking up value of the variable and subtract 1 before passing as argument to the function, do not want to change the value of this variable, so this operator doesn't make sense (if needed it would have to do this operation in another line).#include <iostream> #include <cstdio> using namespace std; long long int factorial(long long int num) { if (num == 1 || num == 0) return 1; return num * factorial(num - 1); } int main() { long long int M, N; while (scanf("%lli %lli", &M, &N) != EOF) cout << factorial(M) + factorial(N) << endl; } See https://ideone.com/UYMeSJ . And https://repl.it/@maniero/Factorial . Also https://github.com/maniero/SOpt/blob/master/CPP/Math/Factorial.cpp .
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Matrix error
Software Programming • c++ liabilities cycles matricians • • Oba22

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int line2=2*line; That's a double number. Great.int** tmpMatrix = new int*[line2]; It's all set aside - it's great!for (int i = 0; i < line; i++) { Oh. Initiate only half, half of the debris index. ♪ ♪ for (int j = 0; j < column; j++) { Although each indicator is initialized column I don't know. ♪ ♪ tmpMatrix[i] = new int[column];
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warning: control reaches end of non-void function [-Wreturn-type] In function ‘char Destination(int)’:
Software Programming • c++ c language warnings • • Avante

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your problem is that in one of the cases the switch will not return anything and the function will not return any value, you have to make all cases return something, you could make the default of the switch return a blank '0' or some character of your preference or when the switch finishes place a return in the same way.char destino (int posicion) { char ciudadDestino; switch (posicion) { case 0: return ciudadDestino = 'A'; break; case 1: return ciudadDestino = 'M'; break; case 2: return ciudadDestino = 'B'; break; case 3: return ciudadDestino = 'V'; break; case 4: return ciudadDestino = 'S'; break; default: return '0'; break; } }
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Examples of indicator in s++
Software Programming • c++ indicators • • Amritpal

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The C+++ language allows access to the addresses of objects in the memory of the programme. The indicator is the variable that contains the object ' s address.Example:int a=10; int *p=&a; In this example, the object is a variable whole with the name "a." This whole variable is worth 10. Also in the program, there's a variable type of "indicator for whole" with the name "p." In this variable, the "p-indicator" is an address for a whole "a."
How to correctly take the C+++ formula. Degree not through pow(s)
Software Programming • c++ competition • • Mystic

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That's example of the code:double series_step(int i, int n, int fact) { if (i > n) return 0; return ((double)i * 2) / ((double)fact + 1) + series_step(i + 1, n, fact * 2); } double series(int n) { return series_step(1, n, 2); } I mean, it's starting. i = 1 and fact = 2 and every terrain we're increasing. i 1 a fact multiplied by 2Programme result n = 2:1.46667
How to make a Java algorithm that can measure runtime of Heap Sort sorting algorithm?
Software Programming • java something • • Bogopo

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You need to first save the initial currentTimeMillis() and then compare with the final.long tempoInicial = System.currentTimeMillis(); // execução do método System.out.println("O método foi executado em " + System.currentTimeMillis() - tempoInicial); System.currentTimeMillis() is a function that returns the difference, measured in milliseconds, between the current time and midnight of January 1, 1970 UTC (coordinated universal time).
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How does C define the variable type?
Software Programming • c language • • jules

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typeid returns the type object http://www.cplusplus.com/reference/std/typeinfo/type_info/ from which type information can be extracted, but it's C+++.
Algorithm with flowchart/diagram
Software Programming • arc languagehive diagram algothm • • Bogopo

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In addition to the problems reported in the first response, this diagram shows how the flow of comparisons would be, where NÃO means that one of the comparisons were not met, therefore valid, and SIM that were met, are not valid.In (k > z) ou (k < y) ou (k > w), that is, respectively:(k > z): the amount of purchase is greater than the maximum quantity;(k < y): the amount of purchase is lower than the minimum quantity, and(k > w): the amount of purchase is greater than the quantity in stock.O Se não x (EOF) ou x (EOF) check whether the product code exists or does not exist.codpro = x; qtdmin = y; qtdmax = z; qtdest = w; qtdcompra = k; Se não x (EOF) ou x (EOF) | | --------------------------------- | (k > z) ou (k < y) ou (k > w) | | | NÃO SIM-------- | | | | [COMPRAR] [NÃO COMPRAR]------ | | [FIM]------------
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Consultation with algorithm for a problem
Software Programming • c++ some • • charlenef

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The problem they want to solve is NOT TSP (which in Spanish is usually called the problem of the trade traveler, not the travel agent, I don't know what they will find if they google the travel agent). I don't know what they call a total solution, I imagine they mean polynomial solution, and yes, effectively TSP does not have known polynomial solutions but DO NOT KNOW if it has polynomial solutions because it is not known if P (the set of problems with polynomial solutions) is not equal to NP (another set of problems, where it is for example TSP).That said, Dijkstra is an algorithm that solves the problem of minimal path, a problem that YES is known to come out in polynomial time.The 10^6 is a lie, in a second there are approximately 10^8 operations, but it is not measured in operations, it is measured in time, it is what takes your program to run, run on your computer with large cases and you will be able to see how long it takes.The problem what you formally ask is given an undirected weighted graph (the city) and some distinguished vertices (the stations) find the least vertex (in index) such that adding to that vertex a station minimizes the maximum distance between u and v for all or vertice of the graph and vertice with station.This is a minimum road problem and goes for example with the Dijkstra algorithm. It also comes with the Floyd-Warshall algorithm that is much easier to implement.If any of these words (grafo, algorithm, computer, vertice, index, distance, Dijkstra, Floyd-Warshall, etc) were not understood, I think they are all in Wikipedia.
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No sign string
Software Programming • c++ char line • • emerson

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Use something like that.using ustring = basic_string<unsigned char>; ustring u; unsigned char c = 240; u = c; cout << int (c);// выводит 240 cout << endl; cout << int(u[0]);//выводит -16, а я хочу 240 It's only gonna be difficult to deal with conventional lines - the same designers will require. unsigned char *there won't be a withdrawal operator. << And so on. ♪ ♪
Algorithm Bower Moura
Software Programming • c language algorithm • • Mystic

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http://www-igm.univ-mlv.fr/~lecroq/string/node14.html Coding the code here so that the answer is not one reference = But it's still worth going.void preBmBc(char *x, int m, int bmBc[]) { int i; for (i = 0; i < ASIZE; ++i) bmBc[i] = m; for (i = 0; i < m - 1; ++i) bmBc[x[i]] = m - i - 1; } void suffixes(char *x, int m, int *suff) { int f, g, i; suff[m - 1] = m; g = m - 1; for (i = m - 2; i >= 0; --i) { if (i > g && suff[i + m - 1 - f] < i - g) suff[i] = suff[i + m - 1 - f]; else { if (i < g) g = i; f = i; while (g >= 0 && x[g] == x[g + m - 1 - f]) --g; suff[i] = f - g; } } } void preBmGs(char *x, int m, int bmGs[]) { int i, j, suff[XSIZE]; suffixes(x, m, suff); for (i = 0; i < m; ++i) bmGs[i] = m; j = 0; for (i = m - 1; i >= 0; --i) if (suff[i] == i + 1) for (; j < m - 1 - i; ++j) if (bmGs[j] == m) bmGs[j] = m - 1 - i; for (i = 0; i <= m - 2; ++i) bmGs[m - 1 - suff[i]] = m - 1 - i; } void BM(char *x, int m, char *y, int n) { int i, j, bmGs[XSIZE], bmBc[ASIZE]; /* Preprocessing */ preBmGs(x, m, bmGs); preBmBc(x, m, bmBc); /* Searching */ j = 0; while (j <= n - m) { for (i = m - 1; i >= 0 && x[i] == y[i + j]; --i); if (i < 0) { OUTPUT(j); j += bmGs[0]; } else j += MAX(bmGs[i], bmBc[y[i + j]] - m + 1 + i); } }
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Testing ofification algorithms
Software Programming • algorithm • • fbio

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Think about how these and other algorithms of sorting will be used for input data (sometimes released) such as:already classified;The retro-reflectored area;Partially organized;Recurring elements;a mass whose values are identical.

Battery algorithms

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Software Programming • c language ponteiro stac languagek • • rielyn

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Software Programming • python time complexity • • carriann

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Software Programming • c++ indicators • • Amritpal

How to correctly take the C+++ formula. Degree not through pow(s)
Software Programming • c++ competition • • Mystic

How to make a Java algorithm that can measure runtime of Heap Sort sorting algorithm?
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Software Programming • c language • • jules

Algorithm with flowchart/diagram
Software Programming • arc languagehive diagram algothm • • Bogopo

Consultation with algorithm for a problem
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No sign string
Software Programming • c++ char line • • emerson

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