The task is to make a letter pyramid. I have done it but its behaviour goes very strange after I pass a certain number of characters that were inputed. Looking forward to an answer.
string input{};
string reverseString{};
getline(cin,input);
for(int j = input.length()-1;j>=0;j--){
reverseString += input.at(j);
}
for(int i = 0;i<input.length();i++){
int numberOfSpaces {};
numberOfSpaces = input.length()-i;
string spaces(" ",numberOfSpaces);
cout<< spaces << input.substr(0,i) << input.at(i)<<reverseString.substr(numberOfSpaces,i) <<spaces<<endl;
}
This is an example of the input/output:
string spaces(" ",numberOfSpaces);
This doesn't do what you think it does.
This constructor takes an array and a count.
It copies count (in this case numberOfSpaces) items from the passed in array.
The passed in array has length 1 (technically 2 with the null terminator), and so anything > 2 will cause undefined behaviour as it reads off the end of the array.
Related
Expected output is "ca" but I'm getting "aca". I have dry ran it, but do not understood why it is doing so. Please, can anyone help me in solving this?
#include<bits/stdc++.h>
using namespace std;
int main()
{
string a = "abbaca";
int i = 0;
while(i < a.size()){
if(a[i] == a[i+1]){
a.erase(i, i+1);
i = 0;
}
else{
i++;
}
}
cout << a;
return 0;
}
a.erase(i, i+1) is wrong.
The string::erase() method takes a starting index and a count, not a pair of indexes denoting a range, as you are thinking.
When removing the duplicate bs, i is 1, and you erase i+1=2 chars, thus "abbaca" becomes "aaca", which is correct. But then, the loop starts over, and when removing the duplicate as, i is 0, so you erase i+1=1 char, thus "aaca" becomes "aca", which is wrong.
You want to remove exact 2 chars each time, so use a.erase(i, 2) instead.
Online Demo
The function 'erase()' erases a part of the string content, shortening the length of the string.The second parameter in the erase function is the count, it means how many characters you want it to remove. If you want 'ca' as expected output you should mention 2 as count of characters to be removed. In the first case it becomes 2 so bb is removed but for 'aa' the count becomes as 1 so 'aca' is the output.
Below is the code for output 'ca', change this erase statement as shown:
if(a[i]==a[i+1]){
a.erase(i,2);
i=0;
}
keep the rest as same
hello i am a beginner in programming and am in the array lessons ,i just know very basics like if conditions and loops and data types , and when i try to solve this problem.
Problem Description
When Serezha was three years old, he was given a set of cards with letters for his birthday. They were arranged into words in the way which formed the boy's mother favorite number in binary notation. Serezha started playing with them immediately and shuffled them because he wasn't yet able to read. His father decided to rearrange them. Help him restore the original number, on condition that it was the maximum possible one.
Input Specification
The first line contains a single integer n (1⩽n⩽105) — the length of the string. The second line contains a string consisting of English lowercase letters: 'z', 'e', 'r', 'o' and 'n'.
It is guaranteed that it is possible to rearrange the letters in such a way that they form a sequence of words, each being either "zero" which corresponds to the digit 00 or "one" which corresponds to the digit 11.
Output Specification
Print the maximum possible number in binary notation. Print binary digits separated by a space. The leading zeroes are allowed.
Sample input:
4
ezor
Output:
0
Sample Input:
10
nznooeeoer
Output:
1 1 0
i got Time limit exceeded on test 10 code forces and that is my code
#include <iostream>
using namespace std;
int main()
{
int n;
char arr[10000];
cin >> n;
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
for (int i = 0; i < n; i++) {
if (arr[i] == 'n') {
cout << "1"
<< " ";
}
}
for (int i = 0; i < n; i++) {
if (arr[i] == 'z') {
cout << "0"
<< " ";
}
}
}
Your problem is a buffer overrun. You put an awful 10K array on the stack, but the problem description says you can have up to 100K characters.
After your array fills up, you start overwriting the stack, including the variable n. This makes you try to read too many characters. When your program gets to the end of the input, it waits forever for more.
Instead of putting an even more awful 100K array on the stack, just count the number of z's and n's as you're reading the input, and don't bother storing the string at all.
According to the compromise (applicable to homework and challenge questions) described here
How do I ask and answer homework questions?
I will hint, without giving a code solution.
In order to fix TLEs you need to be more efficient.
In this case I'd start by getting rid of one of the three loops and of all of the array accesses.
You only need to count two things during input and one output loop.
Im trying to set a char array equal to 2 other arrays depending on if the element in the first array is a number or a letter. The code makes logical sense to me but the output for the 2 other strings after the for loop doesn't correspond to the logic. Is it because of a missing null value somewhere in the other 2 loops or is the code itself invalid? arrayAlpha, arrayNum, and palind are all char arrays set to a length of 30 elements while string length was already determined before the for loop began.
for(int k=0; k<=stringLength; k++)
{
if( isalpha(palind[k])){
arrayAlpha[k]=palind[k];}
if ( isdigit(palind[k]))
{
arrayNum[k]=palind[k];
}
}
Given the input:
char palind[30] = "12345abcde";
arrayAlpha is garbage.
arrayNum is "12345"
However,
char palind[30] = "abcde12345";
arrayAlpha is "abcde".
arrayNum is garbage.
Thus, [k] is the problem when used in your arrayNum or arrayAlpha which doesn't start with 0.
Simple change will just be subtracting the length of the other.
arrayAlpha[k - strlen(arrayNum)] = palind[k];
arrayNum[k - strlen(arrayAlpha)] = palind[k];
since lengthOfPalind = lengthOfArrayAlpha + lengthOfArrayNum assuming palind only contains letters or numbers.
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.
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.