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How do you know the number begins and ends with the same number?

briley last edited by

Can you give me an idea? I'm a little confused with the task: It's natural. Is it true that it begins and ends with the same number? First, I tried to read in the body, then count the elements in the mass to find the first and last number and then compare. But I haven't been able to finish drinking with the mass, and I think it's a bit complicated with the mass.
What logic should be in the decision?
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kalyn

What are you doing with the masses?

last figure - balance of 10 per cent - 10 per cent and final balance

int mynumber=123456;
int last=mynumber%10;
int first=mynumber;
for(first=mynumber;first>=10;first=first/10);        
first = first%10;
printf("%d  - %d",first,last);

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、s2d-x iAd, adMob, iOS, special UIView. So, if you can display the UIView of UIKit in UIs2d-x, it is applied, and then you can write the setting and delegate of iAd and adMob, and the rest should be done by each library. https://stackoverflow.com/questions/19349337/cocos2d-x-admob-on-ios Please try it.
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Program Error: undeclared here (not in a function)
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main.c:16:15: error: 'INT_MIN' undeclared here (not in a function) int fModal = INT_MIN;You need to include <limits.h> at the beginning:#include <limits.h> main.c: In function 'calculo_frequency': main.c:26:16: error: 'false' undeclared (first use in this function) int existe = false;C does not have the word reserved false. It's as easy as doing#include <stdbool.h> This header includes macros true and false.If you don't have it (old C-versions), it's enough that after the includeYou define it. by hand, preferably checking before you're not compiling on C++:#ifndef __cplusplus #define true 1 #define false 0 #endif
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C mass measurement function
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in C++ for masses there is std::vector, as a result not to think of length, it can be obtained through method .size()Well, in C, why do you think the sequence function requires two parameters - a sequence and sequence indicator?because C does not have a function to calculate the arbitrary sequenceExemption - lines that must end '\0' and the size of the line is defined - starting from 0 position of the line indicator is sought '\0' and defined as far as it is, it is the length of the lineIf you look at what it looks like, sizeof(array), then most likely 4 are the size of the indicator, a array[0] - That's it. int And the size of it is also four, and it's the end of one.But if you dochar array[] I'll assume that the size of the mass will be... 4 (4 / 1)
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Apontado C
Software Programming • c language • • obi

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The important part to understand this is to understand how hands work.Let's imagine we declare a pointer with ABC content: const char *c = "ABC";So, "ABC" is somewhere in memory and the pointer is also in memory. If we want to create a pointer for c, it is also possible with the code:const char **cp = &c; In this case the cp pointer is also allocated in memory and is pointing to where the c pointer is.For example, if we have looking at a memory stretch:Endereço 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 A | B | C | | 21 | | 25 | | | Conteúdo cp está no endereço 27 e aponta para o endereço 25. c está no endereço 25 e aponta para o endereço 21. no endereço 21, temos a string ABC.
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You don't have to use regular expressions to accomplish this. You can use string techniques. .startswith() and .endswith():def what_to_do(instructions): if instructions.startswith("Simon says"): return("I " + instructions[10:]) elif instructions.endswith("Simon says"): return("I " + instructions[:-10]) else: return("I won't do it!") print(what_to_do("Simon says jump!")) print(what_to_do("Please, jump!")) Regular expression method:import re r = "(Simon says .)|(. Simon says)" def what_to_do(instructions): if bool(re.fullmatch(r, instructions)): return("I " + re.sub("\sSimon says\s", "", instructions)) else: return("I won't do it!") print(what_to_do("Simon says jump!")) print(what_to_do("Please, jump!")) Explanation of the regular expression:(Simon says .)|(. Simon says)Structure ()|() - the condition is met either in the first bracket or in the second.*. - Any number of any symbols, including underwear.
How do you know the number on the body?
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If this is a standard Tornado template, it should operate from a standard library - range♪ enumerate♪ Somehow:{% for index in range(len(products)) %} <blah index={{ index }}> {{ products[index] }} </blah> {% end %}
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How do you parameterize and transmit the values to SQL, "IN?, ?)" without knowing the number of parameters?
Software Programming • python sql sqlite • • jules

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Try this:sql_pat = """ UPDATE Tasks SET Completed = 1 WHERE User = ? AND task_id IN ({}) """ sql = sql_pat.format(",".join("?" * len(tasks_id))) cursor.execute(sql, [user_id]+ tasks_id) Example:In [6]: user_id = 111; tasks_id = [21, 22, 23] In [7]: sql = sql_pat.format(",".join("?" * len(tasks_id))) In [8]: print(sql) UPDATE Tasks SET Completed = 1 WHERE User = ? AND task_id IN (?,?,?) In [9]: params = [user_id]+ tasks_id In [10]: print(params) [111, 21, 22, 23]
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Cyphorus, end of number
Software Programming • java • • kalyn

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The principle of work is that the last figure will be repeated with a certain frequency, e.g., if the number ends by two, any degree will give the last figure. 2, 4, 8, 6, 2, 4, 8...The algorithms works with the minimum number necessary to meet the challenge, therefore the exceedance MAX_INTEGER He won't.static int lastNumber(int base, int stepen){ if (stepen==0) return 1; int last=base%10; int[] mods={1,1,4,4,2,1,1,4,4,2}; int step=stepen%mods[last]; if (step==0) step=mods[last]; return ((int) Math.pow(last, step))%10; } A little test. Stepeeni won't take more than 17 to avoid exceeding MAX_DOUBLEint count=0; for (int i=0; i<10; i++){ for (int j=0; j<17; j++){ if (lastNumber(i, j)!=(Math.pow(i, j)%10)) count+=1; } } System.out.println(count); // 0
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Create a time limit for the beginning to the end, so the records don't cross.
Software Programming • mysql postgresql database • • fbio

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Postgresql can be used https://postgrespro.ru/docs/postgrespro/9.6/ddl-constraints#ddl-constraints-exclusion (exclusion constraints):CREATE EXTENSION IF NOT EXISTS btree_gist; CREATE TABLE intervals ( type text, publish_up timestamp, publish_down timestamp, EXCLUDE USING GIST (type WITH =, tsrange(publish_up, publish_down) WITH &&) );
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It doesn't detect me in the ascii code on the character I want
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You're running a break When you find something that is not a number, that is, you're coming out of the loop and you stop reading characters. In the file you have spaces, so the program reads 1, then reads a space, and leaves the loop.If you want to read only the numbers, you should have the code this way:#include <stdio.h> #define DIM 1000; int main(){ FILE *file; char caracteres[1000]; file = fopen("texto.txt", "r"); if(file==NULL){ printf("No se puede abrir el archivo\n"); } else{ int i=0; char c; while((c=fgetc(file)) != EOF){ if(c>48 && c<=57){ caracteres[i]=c; printf("%c", caracteres[i]); i++; } } } fclose(file); return 0; } If you look, I've removed the sentence. else so that it does not leave the program and continue with the reading of characters.EDITIf what you want is that you don't take into account the blank spaces, you must keep it in mind in the blank. else:else { if(c==32){ continue; }else{ break; } }
How do you know the number of fields in the structure?
Software Programming • c++ • • Demir

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Well, that's impossible.In C+23, reflexia should be added by which to do so, but until then, it will have to be satisfied with the various questionable hakas.There are several decisions for private cases:If you have all fields in one type of structure, and you know what it is, as proposed by @ARHovsepyan, the size of the structure can be divided into one of its members:int main() { std::cout << sizeof(A) / sizeof(int) << '\n'; std::cout << sizeof(B) / sizeof(float) << '\n'; } Although EMMI, the compilers are allowed to add as many unused baytes to the structure.padding() between and after the fields, in order to remembrance them in a certain way (through this trick would stop working), I have not seen any compiler who would actually do it for the same type of structures.There's another option, more universal.If you haven't created a single designer for your structure, and there's no massiv fields in the structure, you can do this:#include <cstddef> #include <iostream> #include <experimental/type_traits> #include <utility> struct stub { template <typename T> operator T() const; }; template <typename T, typename ...P> using detect_brace_constructible = decltype(T{std::declval<P>()...}); template <typename T, std::size_t ...I> constexpr bool constructible_with_stubs(std::index_sequence<I...>) { return std::experimental::is_detected_v< detect_brace_constructible, T, std::enable_if_t<1 || I, stub>... >; } template <typename T, std::size_t ...I> constexpr int field_count_impl(std::index_sequence<I...>) { int ret = -1; ((void(ret = I), constructible_with_stubs<T>(std::make_index_sequence<I>{})) && ...); return ret-1; } template <typename T, std::size_t MaxFieldCount = 32> inline constexpr int field_count = field_count_impl<T>(std::make_index_sequence<MaxFieldCount+1>{}); struct A { int a,b; }; struct B { float a,b,c; }; int main() { std::cout << field_count<A> << '\n'; // 2 std::cout << field_count<B> << '\n'; // 3 } This template has been continuously testing the structure with an increasing number of arguments:A{} A{stub{}} A{stub{}, stub{}} A{stub{}, stub{}, stub{}} ... Where? stub{} - it's a plug that can be converted into any type.Such construction will be successful, as long as the plugs are as large as the fields in the structure, or less. If the plugs are bigger, there's a mistake. We detect this mistake with SFINAE and return the number of plugs where no mistakes were made.But if there's a body in the structure, it's wrong to consider each element of the body in a separate field of structure.
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Prologue of functions C
Software Programming • c++ c language assembler • • terrea

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The log retains the context of implementation - those data (flags, processor registers, etc.) that may be modified by the function. And the epigologist is rebuilding the context by using the data.The processing of interruptions requires a high speed of work and the code is often written on the assembler (including built-in), and in this case the context can be maintained and rebuilt (or not maintained or rebuilt) without the compiler ' s ability.One more topic: http://msdn.microsoft.com/ru-ru/library/tawsa7cb(v=VS.100).aspx ♪
How do you find the amount, the number of numbers and the lowest number of the text file?
Software Programming • java text • • Analeea

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We need to turn the logic on and figure out what nextInt does. And it's just reading the next number from the file. When the file is finished, there's nothing else to read, and everything but the first one doesn't work. Moreover, the maximum is also incorrect, and the second number is passed.Anyway, all actions have to be done in one passage, read the number, updated the maximum, minimum, the counter, and the amount. int lar = - MaxInt; int min = MaxInt; int count = 0; int sum = 0; while (file.hasnextInt()) { number = file.nextInt(); count++; sum += number; if (number > lar) lar=number; if (number < min) min = number; }
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Indication of structural elements
Software Programming • c language • • Burma

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It's all very simple to do a set of structural and cell shifts. Then all fields can be indexed. The example fills the structure using the cycle.// gcc -Wall -Wextra -Wpedantic -std=c11 offset2.c -o offset2 # include <stddef.h> # include <stdio.h> # include <stdint.h> # include <stdlib.h> typedef struct s_S { int i [ 6 ] ; char c [ 6 ] ; } S ; define indexSize 12 int indexS [ indexSize ] = { offsetof ( S , i [ 0 ] ) , offsetof ( S , i [ 1 ] ) , offsetof ( S , i [ 2 ] ) , offsetof ( S , i [ 3 ] ) , offsetof ( S , i [ 4 ] ) , offsetof ( S , i [ 5 ] ) , offsetof ( S , c [ 0 ] ) , offsetof ( S , c [ 1 ] ) , offsetof ( S , c [ 2 ] ) , offsetof ( S , c [ 3 ] ) , offsetof ( S , c [ 4 ] ) , offsetof ( S , c [ 5 ] ) } ; enum { typeSInt , typeSChar } ; int typeS [ indexSize ] = { typeSInt , typeSInt , typeSInt , typeSInt , typeSInt , typeSInt , typeSChar , typeSChar , typeSChar , typeSChar , typeSChar , typeSChar } ; int main ( ) { volatile S s ; int i ; for ( i = 0 ; i < indexSize ; ++ i ) switch ( typeS [ i ] ) { case typeSInt : * ( volatile int * ) ( ( ( uint8_t * ) & s ) + indexS [ i ] ) = 0 ; break ; case typeSChar : * ( volatile char * ) ( ( ( uint8_t * ) & s ) + indexS [ i ] ) = '0' ; break ; default : fprintf ( stderr , "main : typeS [ %d ] = %d\n" , i , typeS [ i ] ) ; exit ( 1 ) ; } fprintf ( stdout , "s = { { %d , %d , %d , %d , %d , %d } ,\n" " { '%c' , '%c' , '%c' , '%c' , '%c' , '%c' } }\n" , s . i [ 0 ] , s . i [ 1 ] , s . i [ 2 ] , s . i [ 3 ] , s . i [ 4 ] , s . i [ 5 ] , s . c [ 0 ] , s . c [ 1 ] , s . c [ 2 ] , s . c [ 3 ] , s . c [ 4 ] , s . c [ 5 ] ) ; } Type volatile S s ; You need to be because you're changing the structure. Not straight, and the compiler doesn't see you altering the structure yourself.Standard :A pointer to an object type may be converted to a pointer to a different object type. If the resulting pointer is not correctly aligned69) for the referenced type, the behavior is undefined. Otherwise, when converted back again, the result shall compare equal to the original pointer. When a pointer to an object is converted to a pointer to a character type, the result points to the lowest addressed byte of the object. Successive increments of the result, up to the size of the object, yield pointers to the remaining bytes of the object.== sync, corrected by elderman == @elder_manThe indicator on the type of object may be converted into an index to another type of object. If the releasing index is not level enough for the reference type, the conduct is not defined. Otherwise, if reversed, the result would be compared to the reference index. When the indicator on the object is converted to a symbolic type, the result shall indicate the lowest point addressed by the site. Consistent delivery of the result, up to the size of the facility, is provided by the indicators on the remaining site.Used offsetof receive the correct (real) levelling of the structure field.
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histogram of a c++ phrase
Software Programming • c language string histogram • • Burma

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The comparison you have in for: if(frase[num] <= alfabeto[num])az++; Do not count the letters correctly, because it is only comparing if the letter you have is smaller than the letter you go in the alphabet, and it is not in the alphabet.However the error you are getting comes from the last printf in: printf ( "\n\n\tTemons %d letras \"%s\"\n\n" , az, alfabeto[az] //<-- aqui ); For he prints a %s with a letter and not an address. It is important to remember that a type parameter %s Wait a minute char* and not one char, which is the case of alfabeto[az]. You can pass the letter address with &alfabeto[az], but I still don't see the goal of doing that.To count the letters that are part of the alphabet defined by you then need a second for that travels alfabeto and checks whether the current letter matches the alfabeto.I suggest you do so:#include <stdio.h> void histogram(char *); int main() { char str[50]; printf("\n\tDigite uma frase: "); fgets(str,50,stdin); histogram(str); return 0; } void histogram(char *frase) { int num=0, num2=0, az=0; char alfabeto[]="abcdefghijklmnopqrstuvwxyzçABCDEFGHIJKLMNOPQRSTUVWXYZÇ"; for(num = 0; frase[num] != 0; num++) { for (num2 = 0; alfabeto[num2] != '\0'; ++num2){ if (frase[num] == alfabeto[num2]){ az++; break; } } } printf("\n\n\tTemons %d letras\n\n", az); } https://ideone.com/fClxKJ Also note that I changed include for an appropriate C instead of C++.Edit:If you want to know the count of each letter, it is necessary to account for an array, for each font. The very display of the values will have to be done with one for once you have a value for each letter that appears.Example:void histogram(char *frase) { int num=0, num2=0, az=0; char alfabeto[]="abcdefghijklmnopqrstuvwxyzçABCDEFGHIJKLMNOPQRSTUVWXYZÇ"; int contagens[256]; //array para as contagens das letras for (num= 0; num < 256; ++num){ contagens[num]=0; //iniciar o array todo a zeros } for(num = 0; frase[num] != 0; num++) { for (num2 = 0; alfabeto[num2] != '\0'; ++num2){ char letra = frase[num]; if (letra == alfabeto[num2]){ contagens[(int)letra]++; //contabilizar para a letra corrente az++; break; } } } printf("\n\n\tTemons %d letras\n\n", az); for (num = 0; num < 256; ++num){ if (contagens[num] > 0){ //mostrar apenas os que tem pelo menos 1 letra printf("\n%c: %d", (char)num, contagens[num]); } } } https://ideone.com/LLtMNw To start the count array to zeros can alternatively use the function http://www.cplusplus.com/reference/cstring/memset/ , doing memset(contagens, 0, sizeof(contagens));, however it implies including string.h.

How do you know the number begins and ends with the same number?

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