I have got some doubts while solving - Name That Number.
It goes like this -
Among the large Wisconsin cattle ranchers, it is customary to brand cows with serial numbers to please the Accounting Department. The cowhands don't appreciate the advantage of this filing system, though, and wish to call the members of their herd by a pleasing name rather than saying, "C'mon, #4734, get along."
Help the poor cowhands out by writing a program that will translate the brand serial number of a cow into possible names uniquely associated with that serial number. Since the cowhands all have cellular saddle phones these days, use the standard Touch-Tone(R) telephone keypad mapping to get from numbers to letters (except for "Q" and "Z"):
2: A,B,C 5: J,K,L 8: T,U,V
3: D,E,F 6: M,N,O 9: W,X,Y
4: G,H,I 7: P,R,S
Acceptable names for cattle are provided to you in a file named "dict.txt", which contains a list of fewer than 5,000 acceptable cattle names (all letters capitalized). Take a cow's brand number and report which of all the possible words to which that number maps are in the given dictionary which is supplied as dict.txt in the grading environment (and is sorted into ascending order).
For instance, brand number 4734 produces all the following names:
GPDG GPDH GPDI GPEG GPEH GPEI GPFG GPFH GPFI GRDG GRDH GRDI
GREG GREH GREI GRFG GRFH GRFI GSDG GSDH GSDI GSEG GSEH GSEI
GSFG GSFH GSFI HPDG HPDH HPDI HPEG HPEH HPEI HPFG HPFH HPFI
HRDG HRDH HRDI HREG HREH HREI HRFG HRFH HRFI HSDG HSDH HSDI
HSEG HSEH HSEI HSFG HSFH HSFI IPDG IPDH IPDI IPEG IPEH IPEI
IPFG IPFH IPFI IRDG IRDH IRDI IREG IREH IREI IRFG IRFH IRFI
ISDG ISDH ISDI ISEG ISEH ISEI ISFG ISFH ISFI
As it happens, the only one of these 81 names that is in the list of valid names is "GREG".
Write a program that is given the brand number of a cow and prints all the valid names that can be generated from that brand number or ``NONE'' if there are no valid names. Serial numbers can be as many as a dozen digits long.
Here is what I tried to solve this problem. Just go through all the names in the list and check which is satisfying the constraints given.
int numForChar(char c){
if (c=='A'||c=='B'||c=='C') return 2;
else if(c=='D'||c=='E'||c=='F') return 3;
else if(c=='G'||c=='H'||c=='I') return 4;
else if(c=='J'||c=='K'||c=='L') return 5;
else if(c=='M'||c=='N'||c=='O') return 6;
else if(c=='P'||c=='R'||c=='S') return 7;
else if(c=='T'||c=='U'||c=='V') return 8;
else if(c=='W'||c=='X'||c=='Y') return 9;
else return 0;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
freopen("namenum.in","r",stdin);
freopen("namenum.out","w",stdout);
string S; cin >> S;
int len = S.length();
freopen("dict.txt","r",stdin);
string x;
while(cin >> x){
string currName = x;
if(currName.length() != S.length()) continue;
string newString = x;
for(int i=0;i<len;i++){
//now encode the name as a number according to the rules
int num = numForChar(currName[i]);
currName[i] = (char)num;
}
if(currName == S){
cout << newString << "\n";
}
}
return 0;
}
Unfortunately, when I submit it to the judge, for some reason, it says no output produced that is my program created an empty output file. What's possibly going wrong?
Any help would be much appreciated. Thank You.
UPDATE: I tried what Some Programmer Dude suggested by adding a statement else return 0; at the end of the numOfChar function in case of a different alphabet. Unfortunately, it didn't work.
So after looking further at the question and exploring the information for Name That Number. I realized that it is not a current contest, and just a practice challenge. Thus, I updated my answer and also giving you my version of a successful submission. Nonetheless, that is a spoiler and will be posted after why your code was not working.
First, you forgot a } after the declaration of your number function. Secondary, you did not implement anything to check whether if the input fail to yield a valid name. Third, when you use numForChar() on the character of currName, the function yielded an integer value. That is not a problem, the problem is that it is not the ASCII code but is a raw number. You then compare that against a character of the input string. Of which, is an ASCII's value of a digit. Thus, your code can't never find a match. To fix that you can just add 48 to the return value of the numForChar() function or xor the numForChar() return's value to 48.
You are on the right track with your method. But there is a few hints. If you are bored you can always skip to the spoiler. You don't need to use the numForChar() function to actually get a digit value from a character. You can just use a constant array. A constant array is faster than that many if loop.
For example, you know that A, B, C will yield two and A's ASCII code is 65, B's is 66, and C's equal to 67. For that 3, you can have an array of 3 indexes, 0, 1, 2 and all of them stores a 2. Thus, if you get B, you subtract B's ASCII code 65 will yield 1. That that is the index to get the value from.
For getting a number to a character you can have a matrix array of char instead. Skip the first 2 index, 0 and 1. Each first level index, contain 3 arrays of 3 characters that are appropriate to their position.
For dictionary comparing, it is right that we don't need to actually look at the word if the length are unequal. However, besides that, since their dictionary words are sorted, if the word's first letter is lower than the range of the input first letter, we can skip that. On the other hand, if words' first letter are now higher than the highest of the input first letter, there isn't a point in continue searching. Take note that my English for code commenting are almost always bad unless I extensively document it.
Your Code(fixed):
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
int numForChar(char c){
if (c=='A'||c=='B'||c=='C') return 2;
else if(c=='D'||c=='E'||c=='F') return 3;
else if(c=='G'||c=='H'||c=='I') return 4;
else if(c=='J'||c=='K'||c=='L') return 5;
else if(c=='M'||c=='N'||c=='O') return 6;
else if(c=='P'||c=='R'||c=='S') return 7;
else if(c=='T'||c=='U'||c=='V') return 8;
else if(c=='W'||c=='X'||c=='Y') return 9;
else return 0;
}
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
ifstream fin("namenum.in");
ifstream dict("dict.txt");
ofstream fout("namenum.out");
string S;
fin >> S;
int len = S.length();
bool match = false;
string x;
while(dict >> x){
string currName = x;
if(currName.length() != S.length()) continue;
string newString = x;
for(int i=0;i<len;i++){
//now encode the name as a number according to the rules
int num = numForChar(currName[i]) ^ 48;
currName[i] = (char)num;
}
if(currName == S){
fout << newString << "\n";
match = true;
}
}
if ( match == false ){
fout << "NONE" << endl;
}
return 0;
}
Spoiler Code(Improved):
#include <fstream>
#include <string>
using namespace std;
// A = 65
// 65 - 0 = 65
const char wToN[] = {
// A ,B ,C ,D ,E ,F ,G ,H ,I ,
'2','2','2','3','3','3','4','4','4',
// J ,K ,L ,M ,N ,O ,P ,Q ,R ,S
'5','5','5','6','6','6','7','7','7','7',
// T ,U ,V ,W ,X ,Y ,Z
'8','8','8','9','9','9','9'
};
// 2 = {A, B, C} = 2[0] = A, 2[1] = B, 2[2] C
const char nToW[10][3] = {
{}, // 0 skip
{}, // 1
{'A','B','C'},
{'D','E','F'},
{'G','H','I'},
{'J','K','L'},
{'M','N','O'},
{'P','R','S'},
{'T','U','V'},
{'W','X','Y'}
};
int main(){
ifstream fin("namenum.in");
ifstream dict("dict.txt");
ofstream fout("namenum.out");
string S;
fin >> S;
// Since this will not change
// make this a const to make it
// run faster.
const int len = S.length();
// lastlen is last Index of length
// We calculate this value here,
// So we do not have to calculate
// it for every loop.
const int lastLen = len - 1;
int i = 0;
unsigned char digits[len];
unsigned char firstLetter[3];
// If not match print None
bool match = false;
for ( ; i < len; i++ ){
// No need to check upper bound
// constrain did not call for check.
if ( S[i] < '2' ) {
fout << "NONE" << endl;
return 0;
}
}
const char digit1 = S[0] ^ 48;
// There are 3 set of first letter.
// We get them by converting digits[0]'s
// value using the nToW array.
firstLetter[0] = nToW[digit1][0];
firstLetter[1] = nToW[digit1][1];
firstLetter[2] = nToW[digit1][2];
string dictStr;
while(dict >> dictStr){
// For some reason, when keeping the i = 0 here
// it seem to work faster. That could be because of compiler xor.
i = 0;
// If it is higher than our range
// then there is no point contineuing.
if ( dictStr[0] > firstLetter[2] ) break;
// Skip if first character is lower
// than our range. or If they are not equal in length
if ( dictStr[0] < firstLetter[0] || dictStr.length() != len ) continue;
// If we are in the letter range
// we always check the second letter
// not the first, since we skip the first
i = 1;
for ( int j = 1; j < len; j++ ){
// We convert each letter in the word
// to the corresponding int value
// by subtracting the word ASCII value
// to 65 and use it again our wToN array.
// if it does not match the digits at
// this current position we end the loop.
if ( wToN[dictStr[i] - 65] != S[j] ) break;
// if we get here and there isn't an unmatch then it is a match.
if ( j == lastLen ) {
match = true;
fout << dictStr << endl;
break;
}
i++;
}
}
// No match print none.
if ( match == false ){
fout << "NONE" << endl;
}
return 0;
}
I suggest you use c++ file handling. Overwriting stdin and stdout doesn't seem appropriate.
Add these,
std::ifstream dict ("dict.txt");
std::ofstream fout ("namenum.out");
std::ifstream fin ("namenum.in");
Accordingly change,
cin >> S --to--> fin >> S;
cin >> x --to--> dict >> x
cout << newString --to--> fout << newString
I'm new to C and C++ programming, can anyone give me a hint on what I'm doing wrong here. I'm trying to write to concat function that takes to pointers to chars and concatenates the second to the first. The code does do that, but the problem is that it adds a bunch of junk at the end. For instance, when passing the arguments - "green" and "blue", the output will be "greenblue" plus a bunch of random characters. I also wrote the strlen function that strcat uses, which I will provide below it for reference. I'm using the online compiler at https://www.onlinegdb.com/online_c++_compiler
The exact instructions and specification is this:
The strcat(char *__s1, const char *__s2) functions concatenates the contents of __s2 onto __s1 beginning with the NULL character of __s1. Note: The concatenation includes the NULL character of __s2. The function returns __s1.
int main(int argc, char** argv)
{
const int MAX = 100;
char s1[MAX];
char s2[MAX];
cout << "Enter your first string up to 99 characters. ";
cin.getline(s1, sizeof(s1));
int size_s1 = strlen(s1);
cout << "Length of first string is " << size_s1 << "\n";
cout << "Enter your second string up to 99 characters. ";
cin.getline(s2, sizeof(s2));
int size_s2 = strlen(s2);
cout << "Length of second string is " << size_s2 << "\n";
cout << " Now the first string will be concatenated with the second
string ";
char* a = strcat(s1,s2);
for(int i = 0; i<MAX; i++)
cout <<a[i];
// system("pause");
return 0;
}
//strcat function to contatenate two strings
char* strcat(char *__s1, const char *__s2)
{
int indexOfs1 = strlen(__s1);
int s2L = strlen(__s2);
cout <<s2L << "\n";
int indexOfs2 = 0;
do{
__s1[indexOfs1] = __s2[indexOfs2];
indexOfs1++;
indexOfs2++;
}while(indexOfs2 < s2L);
return __s1;
}
//Returns length of char array
size_t strlen(const char *__s)
{
int count = 0;
int i;
for (i = 0; __s[i] != '\0'; i++)
count++;
return (count) / sizeof(__s[0]);
}
The behavior you are seeing is a result of the null terminator of __s1 being overwritten by data from __s2 and no new null terminator being appended. The extra characters you are seeing are just random values in RAM that happen to be after the end of your string. To prevent this a NULL character MUST be added at the end of your string.
A working version is as follows:
char* strcar(char *__s1, const char *__s2)
{
//check inputs for NULL
if(__s1 == NULL || __s2 == NULL)
return NULL;
int s1Length = strlen(__s1);
int s2Length = strlen(__s2);
//ensure strings do not overlap in memory
if(!(__s1 + s1Length < __s2 || __s2 + s2Length < __s1))
return NULL;
//append __s2 to __s1
//the "+ 1" here is necessary to copy the NULL from the end of __s2
for(int i = 0; i < s2Length + 1; i++)
result[s1Length + i] = __s2[i];
}
You Need to add a trailing "\0"-char at the end of __s1.
e.g. insert
__s1[indexOfs1] = 0;
before your return-line.
I recently did this question
Specification:
Input Format The first line contains the number of test cases, T. Next,
T lines follow each containing a long string S.
Output Format For each long string S, display the number of times SUVO
and SUVOJIT appears in it.
I wrote the following code for this :
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int suvo = 0;
int suvojit = 0;
string s;
cin >> s;
for (int i = 0; i <= s.size() - 7; i++) {
if (s.substr(i, 7) == "SUVOJIT")
suvojit++;
}
for (int i = 0; i <= s.size() - 4; i++) {
if (s.substr(i, 4) == "SUVO")
suvo++;
}
cout << "SUVO = " << suvo - suvojit << ", SUVOJIT = " << suvojit << "\n";
}
return 0;
}
The code about gave out of bounds exception for substr() function for this test case:
15
RSUVOYDSUVOJITNSUVOUSUVOJITESUVOSUVOSGSUVOKSUVOJIT
SUVOJITWSUVOSUVOJITTSUVOCKSUVOJITNSUVOSUVOJITSUVOJITSUVOSUVOSUVOJITTSUVOJ
SUVOSUVOSUVOJITASUVOJITGCEBISUVOJITKJSUVORSUVOQCGVHRQLFSUVOOHPFNJTNSUVOJITKSSUVO
SUVOJITSUVOJITJGKSUVOJITISUVOJITKJLUSUVOJITUBSUVOX
MMHBSUVOFSUVOFMSUVOJITUMSUVOJITPSVYBYPMCSUVOJIT
OASUVOSUVOJITSUVOSTDYYJSUVOJITSUVOJITSUVO
RLSUVOCPSUVOJITYSUVOSUVOOGSUVOOESUVOJITMSUVO
WVLFFSUVOJITSUVOVSUVORLESUVOJITPSUVOJITSUVO
RSUVOSUVOJITQWSUVOUMASUVOSUVOJITXNNRRUNUSUVOJIT
HYLSSUVOSUVOSUVOJITPOSUVOJIT
DGMUCSSSUVOJITMJSUVOHSUVOCWTGSUVOJIT
OBNSSUVOYSUVOSUVOJITSUVOJITRHFDSUVODSUVOJITEGSUVOSUVOSUVOJITSUVOSUVOJITSSUVOSUVOSUVOSSUVOJIT
AG
NSUVOJITSUVOSUVOJIT
CGJGDSUVOEASUVOJITSGSUVO
However, when instead of using the s.size() function, I converted the string into a char constant and took the length of it using strlen, then the code caused no error and everything went smoothly.
So, my question is... Why did this happen?
This is my working code with the change:
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int suvo = 0;
int suvojit = 0;
string s;
cin >> s;
int le = strlen(&s[0]);
for (int i = 0; i <= le - 7; i++) {
if (s.substr(i, 7) == "SUVOJIT")
suvojit++;
}
for (int i = 0; i <= le - 4; i++) {
if (s.substr(i, 4) == "SUVO")
suvo++;
}
cout << "SUVO = " << suvo - suvojit << ", SUVOJIT = " << suvojit << "\n";
}
return 0;
}
In one case, you use size_t, in the other case you use int.
If the length is for example 6 characters, then s.size () - 7 is not -1, but one huge number and everything goes wrong. But if you write int len = strlen (...), then len - 7 is indeed -1 and everything is fine.
When I see a number subtracted from size_t, that's an immediate red flag. Write "i + 7 ≤ s.size()", not "i ≤ s.size() - 7".
First of all, in my testing your second leads to a problem as well:
Second, especially with older compilers (well, libraries, really) this can be horrendously inefficient, creating a huge number of temporary strings that you only use to compare with another string1.
So, let's consider how the job should be done instead. std::string has a member named find for situations like this. It returns the position of one string inside another, or std::string::npos if there is none. It allows you to specify a starting position at which to begin searching, when you don't want to start from the beginning.
We also, of course, have two instances of essentially identical code, once to search for SUVO, the other to search for SUVOJIT. The code would be much better off with the search code moved into a function, so we only have the search code in one place.
int count_pos(std::string const &haystack, std::string const &needle) {
size_t pos = 0;
int ret = 0;
while ((pos = haystack.find(needle, pos)) != std::string::npos) {
++ret;
++pos;
}
return ret;
}
Note that this also eliminates quite a bit more messy "stuff" like having to compute the maximum possible position at which at match could take place.
1. Why does compiler/library age matter? Older libraries often used a COW string that dynamically allocated storage for every string. More recent ones typically include what's called a "short string optimization", where storage for a short string is allocated inside the string object itself, avoiding the dynamic allocation.
This little exercise is meant to get a string from the user that could be decimal, hexadecimal, or octal. 1st I need to identify which kind of number the string is. 2nd I need to convert that number to int and display the number in its proper format, eg:
cout <<(dec,hex,oct, etc)<< number;
Here's what I came up with. I'd like a simpler, cleaner way to write this.
string number = "";
cin >> number;
string prefix = "dec";
char zero = '0';
char hex_prefix = 'x';
string temp = "";
int value = 0;
for(int i =0; i<number.size();++i)
{
if(number[0] == zero)//must be octal or hex
{
if (number[0] == zero && number[1] == hex_prefix ) //is hex
{
prefix = "hex";
for(int i = 0; i < (number.size() - 2); ++i)
{
temp[i] = number[i+2];
}
value = atoi(temp.c_str());
}
//... code continues to deal with octal and decimal
You are checking number[0] twice, that's the first most obvious problem.
The inner if already checks both number[0] and number[1], I don't see the point of the outer one.
The outermost loop is also hard to understand, do you expect non-hex data before the number, or what? Your question could be clearer on how the expected input string looks.
I think the cleanest would be to ignore this, and push it into existing (library) code that can parse integers in any base. In C I would recommend strtoul(), you can of course use that in C++ too.
You have two inner loop with same value integer this could be a conflict problem in your code. I suggest you look at the isdigit and islower methods in the c++ library and take advantage of those methods to accomplish your task. isdigit & islower
Good Luck
This is prints the number after deleting the hex prefix, otherwise return 0:
#include<iostream>
#include<cmath>
#include<stdlib.h>
using namespace std;
int main(){
string number = "";
cin >> number;
string prefix = "dec";
char zero = '0';
char hex_prefix = 'x';
string temp = "";
int value = 0;
if (number.size()>=2 && number[0] == zero && number[1] == hex_prefix ) //is hex
{
prefix = "hex";
for(int i = 0; i < (number.size() - 2); ++i)
{
temp[i] = number[i+2];
}
value = atoi(temp.c_str());
}
cout<<value;
return 0;
}
This partial solution that I found is as clean as possible, but it doesn't report the format of the integer:
int string_to_int(std::string str)
{
std::istringstream stream;
stream.unsetf(std::ios_base::dec);
int result;
if (stream >> result)
return result;
else
throw std::runtime_error("blah");
}
...
cout << string_to_int("55") << '\n'; // prints 55
cout << string_to_int("0x37") << '\n'; // prints 55
The point here is stream.unsetf(std::ios_base::dec) - it unsets the "decimal" flag that is set by default. This format flag tells iostreams to expect a decimal integer. If it is not set, iostreams expect the integer in any base.
I'm creating a Caesar Cipher in c++ and i can't figure out how to increment a letter.
I need to increment the letter by 1 each time and return the next letter in the alphabet. Something like the following to add 1 to 'a' and return 'b'.
char letter[] = "a";
cout << letter[0] +1;
This snippet should get you started. letter is a char and not an array of chars nor a string.
The static_cast ensures the result of 'a' + 1 is treated as a char.
> cat caesar.cpp
#include <iostream>
int main()
{
char letter = 'a';
std::cout << static_cast<char>(letter + 1) << std::endl;
}
> g++ caesar.cpp -o caesar
> ./caesar
b
Watch out when you get to 'z' (or 'Z'!) and good luck!
It works as-is, but because the addition promotes the expression to int you want to cast it back to char again so that your IOStream renders it as a character rather than a number:
int main() {
char letter[] = "a";
cout << static_cast<char>(letter[0] + 1);
}
Output: b
Also add wrap-around logic (so that when letter[0] is z, you set to a rather than incrementing), and consider case.
You can use 'a'+((letter - 'a'+n)%26);
assuming after 'z' you need 'a' i.e. 'z'+1='a'
#include <iostream>
using namespace std;
int main()
{
char letter='z';
cout<<(char)('a' + ((letter - 'a' + 1) % 26));
return 0;
}
See this https://stackoverflow.com/a/6171969/8511215
Does letter++ work?
All in all char is a numeric type, so it will increment the ascii code.
But I believe it must be defined as char letter not an array. But beware of adding one to 'Z'. You will get '[' =P
#include <iostream>
int main () {
char a = 'a';
a++;
std::cout << a;
}
This seems to work well ;)
char letter = 'a';
cout << ++letter;
waleed#waleed-P17SM-A:~$ nano Good_morning_encryption.cpp
waleed#waleed-P17SM-A:~$ g++ Good_morning_encryption.cpp -o Good_morning_encryption.out
waleed#waleed-P17SM-A:~$ ./Good_morning_encryption.out
Enter your text:waleed
Encrypted text:
jnyrrq
waleed#waleed-P17SM-A:~$ cat Good_morning_encryption.cpp
#include <iostream>
#include <string>
using namespace std;
int main() {
//the string that holds the user input
string text;
//x for the first counter than makes it keeps looping until it encrypts the user input
//len holds the value (int) of the length of the user input ( including spaces)
int x, len;
//simple console output
cout << "Enter your text:";
//gets the user input ( including spaces and saves it to the variable text
getline(cin, text);
//give the variable len the value of the user input length
len = (int)text.length();
//counter that makes it keep looping until it "encrypts" all of the user input (that's why it keeps looping while its less than len
for(x = 0; x < len; x++) {
//checks each letts (and spaces) in the user input (x is the number of the offset keep in mind that it starts from 0 and for example text[x] if the user input was waleed would be w since its text[0]
if (isalpha(text[x])) {
//converts each letter to small letter ( even though it can be done another way by making the check like this if (text[x] =='z' || text[x] == 'Z')
text[x] = tolower(text[x]);
//another counter that loops 13 times
for (int counter = 0; counter < 13; counter++) {
//it checks if the letts text[x] is z and if it is it will make it a
if (text[x] == 'z') {
text[x] = 'a';
}
//if its not z it will keeps increamenting (using the loop 13 times)
else {
text[x]++;
}
}
}
}
//prints out the final value of text
cout << "Encrypted text:\n" << text << endl;
//return 0 (because the the main function is an int so it must return an integer value
return 0;
}
Note: this is called caeser cipher encryption it works like this :
ABCDEFGHIJKLMNOPQRSTUVWXYZ
NOPQRSTUVWXYZABCDEFGHIJKLM
so for example my name is waleed
it will be written as : JNYRRQ
so its simply add 13 letters to each letter
i hope that helped you
It works but don't forget that if you increment 'z' you need to get 'a' so maybe you should pass by a check function that output 'a' when you get 'z'.
cast letter[n] to byte* and increase its referenced value by 1.