**This is a translation, very hastily written. If you need any clarification, just comment.
The question given is:
We live in a world with too much garbage. We have found a way to compress the garbage, but it can only be done in a specific way, or else the garbage will explode. The garbage has to be laid out in a line, and it can only be compressed with its neighbor, and only if its neighbor has the same value as it.
The first input is int N, and it represents the amount of garbage in the row. The second input is t, and it must have an input of as many characters as the value in N. If the whole thing is able to be compressed until there's only 1 garbage (t) left, then the output will be "YES".
We've figured out that as long as either N == 1, or all inputs in t (all the characters) are the same, the output will be YES.
Example inputs/outputs:
Input:
2
1 1
Output:
YES
Or
Input:
3
1 2 1
Output:
NO
Or
Input:
1
5
Output:
YES
Here's what we've got so far:
#include <iostream>
#include <string>
using namespace std;
int N;
string t;
bool allCharactersSame(string s)
{
int n = s.length();
for (int i = 1; i < n; i++) {
if (s[i] != s[0])
return false;
}
return true;
}
int main()
{
cin>>N;
cin >> t;
if (N == 1)
{
cout << "YES";
}
else if (allCharactersSame(t))
{
cout <<"YES";
}
else
{
cout<<"NO";
}
}
The problem with this is that it outputs YES no matter what, and we think it's because it takes the whitespace of the input into consideration. If we don't include spaces, it works fine. BUT the question dictates that we Have To have spaces separating our inputs. So, we're stumped. Any suggestions?
(I can't comment, therefore I write this as an answer.)
There is some other problem than you think, because the code in the question works as it should. When I gave it input "5 11111" it said "YES" when I gave it "5 12345" it said "NO".
Kai's first comment is slightly weird, when determining whether all characters in a string are the same it is sufficient to compare each of them to the first one, just as you do it in your allCharactersSame() method.
I'd suggest you add some checks on the provided input; the program should probably notice if given N doesn't match given strings' length and it should probably notice when the given string doesn't consist of numbers. As it is now, e.g. input "3 a" says "YES".
Not sure if the title is properly worded, but what I am trying to ask is how would you signify the end of input for an array using newline. Take the following code for example. Not matter how many numbers(more or less) you type during the input for score[6], it must take 6 before you can proceed. Is there a method to change it so that an array can store 6 or 100 variables, but you can decide how many variables actually contain values. The only way I can think of doing this is to somehow incorporate '\n', so that pressing enter once creates a newline and pressing enter again signifies that you don't want to set any more values. Or is something like this not possible?
#include <iostream>
using namespace std;
int main()
{
int i,score[6],max;
cout<<"Enter the scores:"<<endl;
cin>>score[0];
max = score[0];
for(i = 1;i<6;i++)
{
cin>>score[i];
if(score[i]>max)
max = score[i];
}
return 0;
}
To detect "no input was given", you will need to read the input as a input line (string), rather than using cin >> x; - no matter what the type is of x, cin >> x; will skip over "whitespace", such as newlines and spaces.
The trouble with reading the input as lines is that you then have to "parse" the input into numbers. You can use std::stringstream or similar to do this, but it's quite a bit of extra code compared to what you have now.
The typical way to solve this kind of problem, however, is to use a "sentry" value - for example, if your input is always going to be greater or equal to zero, you can use -1 as the sentry. So you enter
1 2 3 4 5 -1
This would reduce the amount of extra code is relatively small - just check if the input is -1, such as
while(cin >> score[i] && score[i] >= 0)
{
...
}
(This will also detect end-of-file, so you could end the input with CTRL-Z or CTRL-D as appropriate for your platform)
I am trying to solve the following code jam question,ive made some progress but for few cases my code give wrong outputs..
Welcome to Code jam
So i stumbled on a solution by dev "rem" from russia.
I've no idea how his/her solution is working correctly.. the code...
const string target = "welcome to code jam";
char buf[1<<20];
int main() {
freopen("input.txt", "rt", stdin);
freopen("output.txt", "wt", stdout);
gets(buf);
FOR(test, 1, atoi(buf)) {
gets(buf);
string s(buf);
int n = size(s);
int k = size(target);
vector<vector<int> > dp(n+1, vector<int>(k+1));
dp[0][0] = 1;
const int mod = 10000;
assert(k == 19);
REP(i, n) REP(j, k+1) {// Whats happening here
dp[i+1][j] = (dp[i+1][j]+dp[i][j])%mod;
if (j < k && s[i] == target[j])
dp[i+1][j+1] = (dp[i+1][j+1]+dp[i][j])%mod;
}
printf("Case #%d: %04d\n", test, dp[n][k]);
}
exit(0);
}//credit rem
Can somebody explain whats happening in the two loops?
Thanks.
What he is doing: dynamic programming, this far you can see too.
He has 2D array and you need to understand what is its semantics.
The fact is that dp[i][j] counts the number of ways he can get a subsequence of the first j letters of welcome to code jam using all the letters in the input string upto the ith index. Both indexes are 1 -based to allow for the case of not taking any letters from the strings.
For example if the input is:
welcome to code jjam
The values of dp in different situations are going to be:
dp[1][1] = 1; // first letter is w. perfect just the goal
dp[1][2] = 0; // no way to have two letters in just one-letter string
dp[2][2] = 1; // again: perfect
dp[1][2] = 1; // here we ignore the e. We just need the w.
dp[7][2] = 2; // two ways to construct we: [we]lcome and [w]elcom[e].
The loop you are specifically asking about calculates new dynamic values based on the already calculated ones.
Whoa, I was practicing this problem few days ago and and stumbled across this question.
I suspect that saying "he's doing dynamic programming" won't not explain too much if you did not study DP.
I can give clearer implementation and easier explanation:
string phrase = "welcome to code jam"; // S
string text; getline(cin, text); // T
vector<int> ob(text.size(), 1);
int ans = 0;
for (int p = 0; p < phrase.size(); ++p) {
ans = 0;
for (int i = 0; i < text.size(); ++i) {
if (text[i] == phrase[p]) ans = (ans + ob[i]) % 10000;
ob[i] = ans;
}
}
cout << setfill('0') << setw(4) << ans << endl;
To solve the problem if S had only one character S[0] we could just count number of its occurrences.
If it had only two characters S[0..1] we see that each occurrence T[i]==S[1] increases answer by the number of occurrences of S[0] before index i.
For three characters S[0..2] each occurrence T[i]==S[2] similarly increases answer by number of occurrences of S[0..1] before index i. This number is the same as the answer value at the moment the previous paragraph had processed T[i].
If there were four characters, the answer would be increasing by number of occurrences of the previous three before each index at which fourth character is found, and so on.
As every other step uses values from the previous ones, this can be solved incrementally. On each step p we need to know number of occurrences of previous substring S[0..p-1] before any index i, which can be kept in array of integers ob of the same length as T. Then the answer goes up by ob[i] whenever we encounter S[p] at i. And to prepare ob for the next step, we also update each ob[i] to be the number of occurrences of S[0..p] instead — i.e. to the current answer value.
By the end the latest answer value (and the last element of ob) contain the number of occurrences of whole S in whole T, and that is the final answer.
Notice that it starts with ob filled with ones. The first step is different from the rest; but counting number of occurrences of S[0] means increasing answer by 1 on each occurrence, which is what all other steps do, except that they increase by ob[i]. So when every ob[i] is initially 1, the first step will run just like all others, using the same code.
Beginners question, on loop efficiency. I've started programming in C++ (my first language) and have been using 'Principles and Practice Using C++' by Bjarne Stroustrup. I've been making my way through the earlier chapters and have just been introduced to the concept of loops.
The first exercise regarding loops asks of me the following:
The character 'b' is char('a'+1), 'c' is char('a'+2), etc. Use a loop to write out
a table of characters with their corresponding integer values:
a 97, b 98, ..., z 122
Although, I used uppercase, I created the following:
int number = 64; //integer value for # sign, character before A
char letter = number;//converts integer to char value
int i = 0;
while (i<=25){
cout << ++letter << "\t" << ++number << endl;
++i;
}
Should I aim for only having 'i' be present in a loop or is it simply not possible when converting between types? I can't really think of any other way the above can be done apart from having the character value being converted to it's integer counterpart(i.e. opposite of current method) or simply not having the conversion at all and have letter store '#'.
Following on from jk you could even use the letter itself in the loop (letter <= 'z'). I'd also use a for loop but that's just me.
for( char letter = 'a'; letter <= 'z'; ++letter )
std::cout << letter << "\t" << static_cast<int>( letter ) << std::endl;
You should aim for clarity first and you try to micro-optimize instead. You could better rewrite that as a for loop:
const int offsetToA = 65;
const int numberOfCharacters = 26;
for( int i = 0; i < numberOfCharacters; ++i ) {
const int characterValue = i + offsetToA;
cout << static_cast<char>( characterValue ) << characterValue << endl;
}
and you can convert between different types - that's called casting (the static_cast construct in the code above).
That's not a bad way to do it, but you can do it with only one loop variable like this:
char letter = 65;
while(letter <= 65+25){
printf("%c\t%d\n", letter, letter);
++letter;
}
there is nothing particularly inefficient about the way you are doing it but it certainly is possible to just convert between chars and ints (a char is an integer type). this would mean you only need to store 1 counter rather than the 3 (i, letter + number) you curently have
also, for looping from a fixed start to end a 'for' loop is perhaps more idiomatic (though its possible you havent met this yet!)
If you are concerned about the efficiency of your loop, I would urge you to try this:
Get this code compiled and running under an IDE, such as Visual Studio, and set a break point at the beginning. When you get there, switch to the disassembly view (instruction view) and start hitting the F11 (single-step) key, and keep a mental count of how many times you are hitting it.
You will see that it enters the loop, compares i against 25, and then starts doing the code for the cout line. That involves incrementing letter, and then going into the << routine for cout. It does a number of things in there, possibly going deeper into subroutines, etc., and finally comes back out, returning an object. Then it pushes "\t" as an argument and passes it to that object, and goes back in and does all the stuff it did before. Then it takes number, increments it, and passes it to the cout::<< routine that accepts an integer, calls a function to convert it to a string (which involves a loop), then does all the stuff it did before to loop that string into the output buffer and return.
Tired? You're not done yet. The endl has to be output, and when that happens, not only does it put "\n" in the buffer, but it calls the system routine to flush that buffer to the file or console where you are sending the I/O. You probably can't F11 into that, but rest assured it takes lots of cycles and doesn't return until the I/O is done.
By now, your F11-count should be in the vicinity of several thousand, more or less.
Finally, you come out and get to the ++i statement, which takes 1 or 2 instructions, and jumps back to the top of the loop to start the next iteration.
NOW, are you still worried about the efficiency of the loop?
There's an easier way to make this point, and it's just as instructive. Wrap an infinite loop around your entire code so it runs forever. While it's running, hit the "pause" button in the IDE, and look at the call stack. (This is called a "stackshot".) If you do this several times you get a good idea of how it spends time. Here's an example:
NTDLL! 7c90e514()
KERNEL32! 7c81cbfe()
KERNEL32! 7c81cc75()
KERNEL32! 7c81cc89()
MSVCRTD! 1021bed3()
MSVCRTD! 1021bd59()
MSVCRTD! 10218833()
MSVCRTD! 1023a500()
std::_Fputc() line 42 + 18 bytes
std::basic_filebuf<char,std::char_traits<char> >::overflow() line 108 + 25 bytes
std::basic_streambuf<char,std::char_traits<char> >::sputc() line 85 + 94 bytes
std::ostreambuf_iterator<char,std::char_traits<char> >::operator=() line 304 + 24 bytes
std::num_put<char,std::ostreambuf_iterator<char,std::char_traits<char> > >::_Putc() line 633 + 32 bytes
std::num_put<char,std::ostreambuf_iterator<char,std::char_traits<char> > >::_Iput() line 615 + 25 bytes
std::num_put<char,std::ostreambuf_iterator<char,std::char_traits<char> > >::do_put() line 481 + 71 bytes
std::num_put<char,std::ostreambuf_iterator<char,std::char_traits<char> > >::put() line 444 + 44 bytes
std::basic_ostream<char,std::char_traits<char> >::operator<<() line 115 + 114 bytes
main() line 43 + 96 bytes
mainCRTStartup() line 338 + 17 bytes
I did this a bunch of times, and not ONCE did it stop in the code for the outer i<=25 loop. So optimizing that loop is like someone's great metaphor: "getting a haircut to lose weight".
Since no one else mentioned it: Having a fixed amount of iterations, this is also a candidate for post-condition iteration with do..while.
char letter = 'a';
do {
std::cout << letter << "\t" << static_cast<int>( letter ) << std::endl;
} while ( ++letter <= 'z' );
However, as shown in Patrick's answer the for idiom is often shorter (in number of lines in this case).
You can promote char to int...
//characters and their corresponding integer values
#include"../../std_lib_facilities.h"
int main()
{
char a = 'a';
while(a<='z'){
cout<<a<<'\t'<<a*1<<'\n'; //a*1 => char operand promoted to integer!
++a;
}
cout<<endl;
}
Incrementing three separate variables is probably a little confusing. Here's a possibility:
for (int i = 0; i != 26; ++i)
{
int chr = 'a' + i;
std::cout << static_cast<char>(chr) << ":\t" << chr << std::endl;
}
Note that using a for loop keeps all the logic of setting up, testing and incrementing the loop variable in one place.
At this point, I wouldn't worry about micro-optimizations such as an efficient way to write a small loop like this. What you have allows a for loop to do the job nicely, but if you are more comfortable with while, you should use that. But I am not sure if that is your question.
I don't think you have understood the question properly. You are writing the code, knowing that 'A' is 65. The whole point of the exercise is to print the value of 'A' to 'Z' on your system, without knowing what value they have.
Now, to get an integer value for a character c, you can do: static_cast<int>(c). I believe that is what you're asking.
I haven't written any code because it should be more fun for you to do so.
Question for the experts: In C, I know that 'a'...'z' need not have continuous values (same for 'A'...'Z'). Is the same true for C++? I would think so, but then it seems highly unlikely that Stroustrup's book assumes that.
thanks for the help.. all i wrote down was
int main()
{
char letter = 96;
int number = letter;
int i = 0;
while(i <26)
{
cout <<++letter <<":" <<++numbers <<" ";
++i;
}
works great...and pretty simple to understand now.
I've tried this and worked fine:
char a = 'a';
int i = a; //represent char a as an int
while (a <= 'z') {
cout << a << '\t' << i << '\n';
++a;
++i;
}
Programming Principles and Practice using C++ (2nd Edition) | Bjarne Stroustrup
Chapter 4 - Computation (Try this #3 - Character Loop)
The character 'b' is char('a'+1), 'c' is char('a'+2), etc. Use
a loop to write out a table of characters with their corresponding integer values:
a 97 b 98 . . . z 122
This is how I solved the problem (from 10 years ago :D)
I am a freshmen btw, so I just started reading this book now... just want to input my solution
#include <iostream>
using namespace std;
int main()
{
int i = 0;
while (i < 26) {
cout << char('a' + i) << '\t' << int(97 + i) << '\n';
++i;
}
}
I solved it by analyzing first the problem which is knowing the char value of 'a' which is 97 up to 'z'. According to this ASCII table
https://www.ascii-code.com/#:~:text=ASCII%20printable%20characters%20%28character%20code%2032-127%29%20Codes%2032-127,digits%2C%20punctuation%20marks%2C%20and%20a%20few%20miscellaneous%20symbols.
Now, we have a clearer understanding on how to solve the said problem.
This is for homework! But I need help anyway. The assignment is to input a sentence then output the number of words, and the number of occurrences of each letter. The output must have the letters in alphabetical order. So far, I've been able to count the number of words and get all the letters to lower case so that I'll be able to keep count of them. My question is how to actually keep count of the letters.
Example of output:
I say Hi.
3 words
1 a
1 h
2 i
1 s
1 y
Here's the code that I have so far:
#include <iostream>
#include <string>
#include <cctype>
using namespace std;
int main()
{
int letters[26];
char letter;
int word = 0;
cout << "Please enter a sentence: "<< endl;
do
{
cin.get(letter);
if(isspace(letter))
word++;
letter = tolower(letter);
cout << letter;
}
while (letter != '\n');
cout << "The number of words = " << word << endl;
return 0;
}
Should I input directly into a C-string? or will that mess up the word count?
If you're allowed to use STL, use std::map for mapping letters to counters. It will additionally sort the letters.
Otherwise, treat chars as indexes in an array of counters and increment them.
My question is how to actually keep
count of the letters
It's fairly straight forward. Simply create an array of 26 integers, (one for each letter), and initialize it to zero.
int letters[26] = { 0 }; // Initialize array to zero
Each value in the array corresponds to a count of a particular letter. Array index 0 refers to 'a', array index 1 refers to 'b', and so on. Then, everytime you encounter a letter, increment the appropriate value in the array. You can use the character 'a' (ASCII value 97) as a starting offset. So, given the variable char letter; you would do:
++letters[tolower(letter) - 'a'];
But always make sure that before you increment the appropriate value in the array, you check that isalpha(letter) && islower(letter) to make sure that your letter is in the range of lowercase a-z; otherwise you will access an index beyond the bounds of the array. You can also test for this condition by saying if (letter >= 'a' && letter <= 'z').
Hint: tolower(letter)-'a' is:
0 if letter is a
1 if letter is b
...
Hm, just few points to make your home task more useful to you (and your code more correct):
Think what happens if you have file with several spaces in a row (word counting).
Think how to be more correct with 'letters' (check for isalpha() at least). Also isalpha() could be key for simpler counting with fixed array [256] (this might be even the best solution as for performance vs std::map usage, check std::map documentation anyway).
Think about more effective file input. At least line at once.