I'm trying to solve this problem from an online judge (Codeforces):
One day Deivis came across two Vectors of integers A and B, and wondered, could it be possible to form the number X by adding an element of A to another element of B?
More formally, it is possible to choose two indexes i and j such that Ai + Bj = x?
Input
The first entry line is two integers n and x. The second line contains n numbers, the vector A. The third and last line contains n numbers, vector B.
Output
Print 1 if it is possible to form the number x from a sum of one element of each vector, and 0 otherwise."
My problem is that I can not fill in the second vector, when the program runs on the site it fills the vector with zeros. I am using C ++, here's my code:
#include <bits/stdc++.h>
using namespace std;
#define MAX 10
int main()
{
int n, x, i = 0, j = 0, resp = 0, sum;
vector<int> vetA(MAX), vetB(MAX);
cin >> n >> x;
while (scanf("%d", &vetA[i]) == 1)
i++;
while (scanf("%d", &vetB[j]) == 1)
j++;
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
sum = vetA[i] + vetB[j];
if (sum == x)
{
resp = 1;
goto END;
}
}
}
END: printf("%d", resp);
return 0;
}
I try to use getchar() after each while loop, but seems that on the site it does not do data capture like on a keyboard, and so the second vector isn't receiving any data. I've also tried to capture data as a std::string but that doesn't work.
Can someone help me?
Here are some hints/examples to compare your program to:
#include <iostream> //Include each standard library seperately
#include <vector> //#include <bits/stdc++.h> is bad practice
// Only declare variables as they are used.
int n; // Better coding practice is one variable per line.
int x; // Competitions shouldn't care how many lines.
if (!(std::cin >> n >> x)) //This is basically the same as cin.fail()
{
std::cerr << "Error inputting data.\n";
return 1;
}
// Now create the vectors, after the size has read in.
std::vector<int> vetA(n);
std::vector<int> vetB(n);
// The number of elements is known, so use a "for" loop.
for (size_t i = 0; i < n; ++i)
{
std::cin >> vetA[i];
}
for (size_t i = 0; i < x; ++i)
{
std::cin >> vetB[i];
}
You should add in some error handling because your program will be given some invalid inputs.
The inputs and vector sizes are examples since you didn't specify the input format in your Post.
Related
I'm stuck for the first time on a lab for this class. Please help!
The prompt is:
Write a program that reads a list of integers, and outputs those integers in reverse. The input begins with an integer indicating the number of integers that follow. For coding simplicity, follow each output integer by a comma, including the last one.
Ex: If the input is:
5 2 4 6 8 10
the output is:
10,8,6,4,2,
2 questions: (1) Why does the vector not take user input unless the const int is included? (2) Why does the code not work in general? It seems to properly output, but with an error, and does not include the end line?
#include <iostream>
#include <vector>
using namespace std;
int main() {
const int MAX_ELEMENTS = 20;
vector<int> userInts(MAX_ELEMENTS);
unsigned int i;
int numInts;
cin >> numInts;
for (i = 0; i < numInts; ++i) {
cin >> userInts.at(i);
}
for (i = (numInts - 1); i >= 0; --i) {
cout << userInts.at(i) << ",";
}
cout << endl;
return 0;
}
Firstly, you need to specify the size because you are not using the vector's push_back functionality. Since you are only using at, you must specify the size ahead of time. Now, there's a few ways to do this.
Example 1:
cin >> numInts;
vector<int> userInts(numInts); // set the size AFTER the user specifies it
for (i = 0; i < numInts; ++i) {
cin >> userInts.at(i);
}
Alternatively, using push_back you can do:
vector<int> userInts; // set the size AFTER the user specifies it
for (i = 0; i < numInts; ++i) {
int t;
cin >> t;
userInts.push_back(t);
}
As for looping backwards, i >= 0 will always be true for unsigned numbers. Instead, you can use iterators.
for ( auto itr = userInts.rbegin(); itr != userInts.rend(); ++itr ) {
cout << *itr;
}
If you need to use indexes for the reverse loop, you can do:
for ( i = numInts - 1; i != ~0; --i ) { // ~0 means "not 0", and is the maximum value, I believe this requires c++17 or 20 though
cout << userInts.at(i);
}
with unsigned int i; the condition i >= 0 is always true. Eventually you will access an out-of-range element, which will throw std::out_of_range.
To answer your other question
std::vector userInts;
create a vector with no entries
userInts.at(i)
tries to access the (non existnat) ith entry.
You have 2 choices
create vector with a lot of empty etries
ask the vector to dynamically grow
The first one is what you did
const int MAX_ELEMENTS = 20;
vector<int> userInts(MAX_ELEMENTS);
Or you can do
userInts.push_back(x);
this will make sure there is enough space in the vector and add the new element to the end.
The problem is the following: We are given a number 's', s ∈ [0, 10^6], and a number 'n', n ∈ [0, 50000], then n numbers, and we have to find how many number pairs' sum is equal to the 's' number (and we must use either maps or sets to solve it)
Here is the example:
Input:
5 (this is s)
6 (this is n)
1
4
3
6
-1
5
Output:
2
explanation : these are the (1,4) and (6,−1) pairs. (1 +4 = 5 and 6 + (-1) = 5)
Here is my "solution" , I don't even know if it's correct, but it works for the example that we got.
#include <iostream>
#include <map>
#include <iterator>
using namespace std;
int main()
{
cin.tie(0);
ios::sync_with_stdio(false);
int s;
cin >> s;
int n;
cin >> n;
map<int, int> numbers;
int element;
int counter = 0;
for(int i=0; i<n;i++)
{
cin >> element;
numbers.insert(pair<int, int>(element, s-element));
}
for(map<int, int>::iterator it = numbers.begin(); it != numbers.end(); it++)
{
map<int, int>::iterator it2 = it;
while(it2 != numbers.end())
{
if(it->second == it2->first)
{
counter++;
break;
}
it2++;
}
}
cout << counter << "\n";
return 0;
}
Thanks for the answers in advance! I'm still a beginner and I'm learning, sorry.
element, s-element is a good idea but there is no reason to store all the pairs and only then check for duplicates. This removes the O(n^2) loop you have there at the end.
The standard way using hashing would be:
seen=unordered_map<number,count>()
for 1...n:
e = read_int()
if (s-e) in seen:
duplicates+=seen[s-e] # Found new seen[s-e] duplicates.
if e in seen:
seen[e]+=1
else:
seen.insert(e,1)
return duplicates
Here's a brute-force method, using a vector:
int target_s = 0;
int quantity_numbers = 0;
std::cin >> target_s >> quantity_numbers;
std::vector<int> data(quantity_numbers);
for (int i = 0; i < quantity_numbers; ++i)
{
cin >> data[i];
}
int count = 0;
for (int i = 0; i < quantity_numbers; ++i)
{
for (j = 0; j < quantity_numbers; ++j)
{
if (i == j) continue;
int pair_sum = data[i] + data[j];
if (pair_sum == target_s) ++count;
}
}
std::cout << count;
The above code includes the cases where pair <a,b> == s and pair <b,a> == s. Not sure if the requirement only wants pair <a,b> in this case.
As always with this kind of questions, the selection of the appropriate algorithm will improve your solution. Writing some "better" C++ code, will nearly never help. Also, brute forcing is nearly never a solution for such an algorithm.
With the following described approach (which was of course not invented by me), we need just one std::map (or even better, a std::unordered_map) and one for loop. We do not need to store the read values in an additional std::vector or such alike. So, we can come up with low memory condumption and fast computation.
Approach. Any time, after reading a value, we will calculate the delta from the desired sum.
If we look at the required condition that the current value and some previuosly read value, should add up to the desired sum, we can write the following mathematical equations:
currentValue + previouslyReadValue = desiredSum
or
desiredSum - currentValue = previouslyReadValue
or with
delta = desiredSum - currentValue
-->
delta == previouslyReadValue
So, we need to look at the already read values and if they are equal to the delta (Because then they would add up the the desired sum), add their count of occurence the the resulting count of valid pairs.
The already read values and their count of occurence will be stored in a std::unordered_map.
All this will result in a 10 line solution:
#include <iostream>
#include <unordered_map>
int main() {
// Initialize our working variables
int numberOfValues{}, desiredSum{}, currentValue{}, resultingCount{};
// Read basic parameters. Desired sum and overall number of input values.
std::cin >> desiredSum >> numberOfValues;
// Here, we will store all values and their count of occurence
std::unordered_map<int, int> valuesAndCount{};
// Read all values and operate on them
for (int i{}; i < numberOfValues; ++i) {
std::cin >> currentValue; // Read from cin
const int delta{ desiredSum - currentValue }; // Calculate the delta from the desired sum
// Look, if the calculated delta is already in the map. Becuase, if the delta and the
// current value sum up to our desired sum, then we found a valid pair.
if (valuesAndCount.find(delta) != valuesAndCount.end())
// Increase the resulting count, by the number of times that this delta value has already been there
resultingCount += valuesAndCount[delta];
// Nothing special, Just cound the occurence of this value.
valuesAndCount[currentValue]++;
}
return !!(std::cout << resultingCount);
}
This question already has answers here:
Converting integer into array of digits [closed]
(4 answers)
Closed 2 years ago.
I know there are similar questions on Stack overflow about this. I already checked them. Here are two points:
The number will be an input from the user, so I won't know how many digits the number may actually contain
I DON'T want to directly print the digits, I need to do some further process on every digit, so I need a way to save or assign every single digit.
So fore example, if the user enters 1027, I'll need 1, 0, 2, 7 returned, and not printed. That's where my problem starts.
Here's how I'd write it if it could be printed:
int x;
cin>>x;
while(x != 0)
{
cout<<x%10<<endl;
x/=10;
}
Any hints or help is in advance appreciated.
It depends what order you need it in. If you need least-significant digit (rightmost) to most-significant (leftmost), then your solution is almost there
int x = ...
while(x != 0)
{
int current = x % 10; // get rightmost digit
x /= 10;
// process 'current', or store in a container for later processing
}
If you need most-significant (leftmost) to least-significant (rightmost), then you can do this recursively:
void for_each_digit(int input)
{
// recursive base case
if (input == 0) { return; };
int current = input % 10
for_each_digit(input / 10); // recurse *first*, then process
// process 'current', add to container, etc
}
// ...
int x = ...
for_each_digit(x);
Edit: I apparently missed the part about returning a sequence of digits.
Either approach works. If you go right-to-left, you will need to reverse the container first. If you go recursive, you will need to append each value to the container.
Use a std::string:
std::string input;
std::cin >> input;
Now input[i] is the i-th digit. input.size() is the number of digits.
Well you can use vector. It can take variable length input. You need not declare the size beforehand. Learn more about vector here: vector
#include<iostream>
#include<vector>
#include <algorithm> // std::reverse
using namespace std;
int main(void)
{
vector<int>digits;
int x;
cin >> x;
while(x)
{
digits.push_back(x % 10);
x = x / 10;
}
// reversing the order of the elements inside vector "digits" as they are collected from last to first and we want them from first to last.
reverse(digits.begin(), digits.end());
// Now the vector "digits" contains the digits of the given number. You can access the elements of a vector using their indices in the same way you access the elements of an array.
for(int i = 0; i < digits.size(); i++) cout << digits[i] << " ";
return 0;
}
You may try std::vector<int> to store unknown number of integers as shown:
#include <iostream>
#include <vector>
int main(void) {
std::vector<int> digits;
std::string s;
std::cout << "Enter the number: ";
std::cin >> s;
size_t len = s.length();
for (size_t i = 0; i < len; i++) {
digits.push_back(s[i] - '0');
}
// Comment next 3 code to stop getting output
for (size_t i = 0; i < len; i++)
std::cout << digits[i] << ' ';
std::cout << std::endl;
return 0;
}
Note: This approach doesn't performs any mathematical operations (i.e. operation of division and getting remainders). It simply stores each integer in a vector using a for loop.
I'm trying to take in some input and find the number of a certain character in a string. I keep getting a weird answer when I try to take in the actual string. Why is this happening?
I'm using cout to find why I'm getting such weird numbers and it appears to be a problem with the input.
Note - This is my attempted solution to Codeforces Problem 462 B. I'm attempting to just find the number of a certain letter in the input. My friend is attempting a bubble sort method.
Input:
6 4
YJSNPI
Expected Output:
YJSNPI
4
Actual Output:
YJSNPI
1699623981
Code:
#include <iostream>
#include <string>
#include <vector>
#include <istream>
using namespace std;
int main()
{
int n, k, counting;
cin >> n >>k;
char trash;
cin.get(trash);
vector<string> cards;
string theline, name;
cin >> theline;
cout << theline << "\n";
for (int i = 0; i < n; i++){
name = theline[i];
cards.push_back(name);
}
for (int i = 0; i < n; i++){
if (cards[i] == cards[k-1]){
counting++;
}
}
int tmp = 0;
if (cards.size() != k){
tmp = k - counting;
}
counting *= k;
counting += tmp;
cout << counting;
return 0;
}
Local variables are not automatically initialized to 0. If you try to use the value of a local variable before assigning it, you get undefined behavior. You're incrementing counting without ever initializing it. Change to:
int n, k, counting = 0;
The issue is that the variable "counting" is never initialized - handy link
Basically, "counting" has some garbage value from memory after you declare it with
int counting;
Then, the first operation performed is
counting++;
And the garbage value is saved.
THE FIX:
Change
int counting;
to
int counting = 0;
NOTE: n and k are not helpful variable names. It would make understanding the code a lot easier if they had real names, but oh well.
ADDITIONALLY:
As chris mentioned above, make the compiler work for you. See comment below for good compiler flags. Don't ignore warnings!
Can't really understand what you are doing here. But I can see where you are going wrong.
int n, k, counting;
counting is uninitialized try
int n, k, counting = 0;
I get answer of (1*4 + 4 - 1) = 7 not the 4 you are expecting.
This code will always result in counting = 1, given that k is within range.
for (int i = 0; i < n; i++){
if (cards[i] == cards[k-1]){
counting++;
}
}
https://ideone.com/7Hk3ix
I have to find all possible, unique substrings from a bunch of user-input strings. This group of substrings has to be alphabetically sorted without any duplicate elements, and the group must be queryable by number. Here's some example input and output:
Input:
3 // This is the user's desired number of strings
abc // So the user inputs 3 strings
abd
def
2 // This is the user's desired number of queries
7 // So the user inputs 2 queries
2
Output:
// From the alphabetically sorted group of unique substrings,
bd // This is the 7th substring
ab // And this is the 2nd substring
Here's my implementation:
#include <map>
#include <iostream>
using namespace std;
int main() {
int number_of_strings;
int number_of_queries;
int counter;
string current_string;
string current_substr;
map<string, string> substrings;
map<int, string> numbered_substrings;
int i;
int j;
int k;
// input step
cin >> number_of_strings;
string strings[number_of_strings];
for (i = 0; i < number_of_strings; ++i)
cin >> strings[i];
cin >> number_of_queries;
int queries[number_of_queries];
for (i = 0; i < number_of_queries; ++i)
cin >> queries[i];
// for each string in 'strings', I want to insert every possible
// substring from that string into my 'substrings' map.
for (i = 0; i < number_of_strings; ++i) {
current_string = strings[i];
for (j = 1; j <= current_string.length(); ++j) {
for (k = 0; k <= current_string.length()-j; ++k) {
current_substr = current_string.substr(k, j);
substrings[current_substr] = current_substr;
}
}
}
// my 'substrings' container is now sorted alphabetically and does
// not contain duplicate elements, because the container is a map.
// but I want to make the map queryable by number, so I'm iterating
// through 'substrings' and assigning each value to an int key.
counter = 1;
for (map<string,string>::iterator it = substrings.begin();
it != substrings.end(); ++it) {
numbered_substrings[counter] = it->second;
++counter;
}
// output step
for (i = 0; i < number_of_queries; ++i) {
if (queries[i] > 0 && queries[i] <= numbered_substrings.size()) {
cout << numbered_substrings[queries[i]] << endl;
} else {
cout << "INVALID" << endl;
}
}
return 0;
}
I need to optimize my algorithm, but I'm not sure how to do it. Maybe it's the fact that I have a second for loop for assigning new int keys to each substring. Help?
Check out Suffix tree. It usually runs in O(n) time:
This article was helpful for me:
http://allisons.org/ll/AlgDS/Tree/Suffix/
Minor notes:
1. include <string>
2. careful with those } else {; one day you'll have a lot of else if branches
and a lot of lines and you'll wonder where an if starts and where it ends
3. careful with unsigned versus signed mismatching... again, one day it will
come back and bite (also, it's nice to compile without errors or warnings)
4. don't try to define static arrays with a variable size
5. nice with ++ i. not many know it has a slight performance boost
(maybe not noticeable with today's processors but still)
While I do agree that using proper algorithms when needed (say bubble sort, heap sort etc. for sorting, binary search, binary trees etc. for searching), sometimes I find it nice to do an optimization on current code. Imagine having a big project and implementing something requires rewrites... not many are willing to wait for you (not to mention the required unit testing, fat testing and maybe fit testing). At least my opinion. [and yes, I know some are gonna say that if it is so complicated then it was written badly from the start - but hey, you can't argue with programmers that left before you joined the team :P]
But I do agree, using existing stuff is a good alternative when called for. But back to the point. I tested it with
3, abc, def, ghi
4, 1, 3, 7, 12
I can't say whether yours is any slower than mine or vice-versa; perhaps a random string generator that adds maybe 500 inputs (then calculates all subs) might be a better test, but I am too lazy at 2 in the morning. At most, my way of writing it might help you (at least to me it seems simpler and uses less loops and assignments). Not a fan of vectors, cos of the slight overhead, but I used it to keep up with your requirement of dynamic querying... a static array of a const would be faster, obviously.
Also, while not my style of naming conventions, I decided to use your names so you can follow the code easier.
Anyway, take a look and tell me what you think:
#include <map>
#include <iostream>
#include <string> // you forgot to add this... trust me, it's important :)
#include <vector> // not a fan, but it's not that bad IF you want dynamic buffers
#include <strstream>
using namespace std;
int main ()
{
unsigned int number_of_strings = 0;
// string strings[number_of_strings]; // don't do this... you can't assign static arrays of a variable size
// this just defaults to 0; you're telling the compiler
cin >> number_of_strings;
map <string, string> substrings;
string current_string, current_substr;
unsigned int i, j, k;
for (i = 0; i < number_of_strings; ++ i)
{
cin >> current_string;
substrings[current_string] = current_string;
for (j = 1; j <= current_string.length(); ++ j)
{
for (k = 0; k <= current_string.length() - j; ++ k)
{
current_substr = current_string.substr(k, j);
substrings[current_substr] = current_substr;
}
}
}
vector <string> numbered_substrings;
for (map <string, string>::iterator it = substrings.begin(); it != substrings.end(); ++ it)
numbered_substrings.push_back(it->second);
unsigned int number_of_queries = 0;
unsigned int query = 0;
cin >> number_of_queries;
current_string.clear();
for (i = 0; i < number_of_queries; ++ i)
{
cin >> query;
-- query;
if ((query >= 0) && (query < numbered_substrings.size()))
current_string = current_string + numbered_substrings[query] + '\n';
else
cout << "INVALID: " << query << '\n' << endl;
}
cout << current_string;
return 0;
}