For this program a user must enter 10 contestants and the amount of second it took for them to complete a swimming race. My problem is that I must output the 1st, 2nd and 3rd placers, so I need to get the three smallest arrays (as they would be the quickest times) but I'm unsure on how to do it. Here is my code so far.
string names[10] = {};
int times[10] = { 0 };
int num[10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int min1 = 0, min2 = 0, min3 = 0;
cout << "\n\n\tCrawl";
for (int i = 0; i < 10; i++)
{
cout << "\n\n\tPlease enter the name of contestant number " << num[i] << ": ";
cin >> names[i];
cout << "\n\tPlease enter the time it took for them to complete the Crawl style: ";
cin >> times[i];
while (!cin)
{
cout << "\n\tError! Please enter a valid time: ";
cin.clear();
cin.ignore();
cin >> times[i];
}
if (times[i] < times[min1])
min1 = i;
cout << "\n\n\t----------------------------------------------------------------------";
}
system("cls");
cout << "\n\n\tThe top three winners of the Crawl style race are as follows";
cout << "\n\n\t1st Place - " << names[min1];
cout << "\n\n\t2nd Place - " << names[min2];
cout << "\n\n\t3rd Place - " << names[min3];
}
_getch();
return 0;
}
As you can see, it is incomplete. I know how to get the smallest number, but its the second and third smallest that is giving me trouble.
your code is full of errors:
what do you do with min2 and min3 as long as you don't assign them?? they are always 0
try checking: cout << min2 << " " << min3;
also you don't initialize an array of strings like that.
why you use an array of integers for just printing number of input:
num? instead you can use i inside loop adding to it 1 each time
to solve your problem use a good way so consider using structs/clusses:
struct Athlete
{
std::string name;
int time;
};
int main()
{
Athlete theAthletes[10];
for(int i(0); i < 10; i++)
{
std::cout << "name: ";
std::getline(std::cin, theAthletes[i].name);
std::cin.sync(); // flushing the input buffer
std::cout << "time: ";
std::cin >> theAthletes[i].time;
std::cin.sync(); // flushing the input buffer
}
// sorting athletes by smaller time
for(i = 0; i < 10; i++)
for(int j(i + 1); j < 10; j++)
if(theAthletes[i].time > theAthletes[j].time)
{
Athlete tmp = theAthletes[i];
theAthletes[i] = theAthletes[j];
theAthletes[j] = tmp;
}
// printing the first three athletes
std::cout << "the first three athelets:\n\n";
std::cout << theAthletes[0].name << " : " << theAthletes[0].time << std::endl;
std::cout << theAthletes[1].name << " : " << theAthletes[1].time << std::endl;
std::cout << theAthletes[2].name << " : " << theAthletes[2].time << std::endl;
return 0;
}
I hope this will give u the expected output. But i suggest u to use some sorting alogirthms like bubble sort,quick sort etc.
#include <iostream>
#include<string>
using namespace std;
int main() {
int times[10] = { 0 };
int num[10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int min1 = 0, min2 = 0, min3 = 0,m;
string names[10] ;
cout << "\n\n\tCrawl";
for (int i = 0; i < 10; i++)
{
cout << "\n\n\tPlease enter the name of contestant number " << num[i] << ": ";
cin >> names[i];
cout << names[i];
cout << "\n\tPlease enter the time it took for them to complete the Crawl style: ";
cin >> times[i];
cout<<times[i];
while (!cin)
{
cout << "\n\tError! Please enter a valid time: ";
cin.clear();
cin.ignore();
cin >> times[i];
}
if(times[i]==times[min1]){
if(times[min1]==times[min2]){
min3=i;
}else{min2 =i;}
}else if(times[i]==times[min2]){
min3=i;
}
if (times[i] < times[min1]){
min1 = i;
cout <<i;
}
int j=0;
while(j<i){
if((times[j]>times[min1])&&(times[j]<times[min2])){
min2 =j;
j++;
}
j++;
}
m=0;
while(m<i){
if((times[m]>times[min2])&&(times[m]<times[min3])){
min3 =m;
m++;
}
m++;
}
cout << "\n\n\t----------------------------------------------------------------------";
}
cout << "\n\n\tThe top three winners of the Crawl style race are as follows";
cout << "\n\n\t1st Place - " << names[min1];
cout << "\n\n\t2nd Place - " << names[min2];
cout << "\n\n\t3rd Place - " << names[min3];
return 0;
}
There is actually an algorithm in the standard library that does exactly what you need: std::partial_sort. Like others have pointed out before, to use it you need to put all the participant data into a single struct, though.
So start by defining a struct that contains all relevant data. Since it seems to me that you only use the number of the contestants in order to be able to later find the name to the swimmer with the fastest time, I'd get rid of it. Of course you could also add it back in if you like.
struct Swimmer {
int time;
std::string name;
};
Since you know that there always will be exactly 10 participants in a race, you can also go ahead and replace the C-style array by a std::array.
The code to read in the users then could look like this:
std::array<Swimmer, 10> participants;
for (auto& participant : participants) {
std::cout << "\n\n\tPlease enter the name of the next contestant: ";
std::cin >> participant.name;
std::cout << "\n\tPlease enter the time it took for them to complete the Crawl style: ";
while(true) {
if (std::cin >> participant.time) {
break;
}
std::cout << "\n\tError! Please enter a valid time: ";
std::cin.clear();
std::cin.ignore();
}
std::cout << "\n\n\t----------------------------------------------------------------------";
}
Partial sorting is now essentially a one-liner:
std::partial_sort(std::begin(participants),
std::begin(participants) + 3,
std::end(participants),
[] (auto const& p1, auto const& p2) { return p1.time < p2.time; });
Finally you can simply output the names of the first three participants in the array:
std::cout << "\n\n\tThe top three winners of the Crawl style race are as follows";
std::cout << "\n\n\t1st Place - " << participants[0].name;
std::cout << "\n\n\t2nd Place - " << participants[1].name;
std::cout << "\n\n\t3rd Place - " << participants[2].name << std::endl;
The full working code can be found on coliru.
This is not a full solution to your problem, but just meant to point you into the right direction...
#include <iostream>
#include <limits>
#include <algorithm>
using namespace std;
template <int N>
struct RememberNsmallest {
int a[N];
RememberNsmallest() { std::fill_n(a,N,std::numeric_limits<int>::max()); }
void operator()(int x){
int smallerThan = -1;
for (int i=0;i<N;i++){
if (x < a[i]) { smallerThan = i; break;}
}
if (smallerThan == -1) return;
for (int i=N-1;i>smallerThan;i--){ a[i] = a[i-1]; }
a[smallerThan] = x;
}
};
int main() {
int a[] = { 3, 5, 123, 0 ,-123, 1000};
RememberNsmallest<3> rns;
rns = std::for_each(a,a+6,rns);
std::cout << rns.a[0] << " " << rns.a[1] << " " << rns.a[2] << std::endl;
// your code goes here
return 0;
}
This will print
-123 0 3
As you need to know also the names for the best times, you should use a
struct TimeAndName {
int time;
std::string name;
}
And change the above functor to take a TimeAndName instead of the int and make it also remember the names... or come up with a different solution ;), but in any case you should use a struct similar to TimeAndName.
As your array is rather small, you could even consider to use a std::vector<TimeAndName> and sort it via std::sort by using your custom TimeAndName::operator<.
Related
I have only a little experience in C++. I'm trying to write a program to print each element of the 'sales' array:
#include <iostream>
#include <iomanip>
using namespace std;
void printArray(int, int);
int main()
{
char chips[5][50] = {"mild", "medium", "sweet", "hot", "zesty"};
int sales[5] = {0};
int tempSales, counter;
const int i = 5;
for (counter = 0; counter < i; counter++)
{
cout << "Please enter in the sales for " << chips[counter] << ": ";
cin >> tempSales;
tempSales >> sales[counter][5];
}
cout << "{";
for (int counter = 0; counter < i; counter++)
{
cout << chips[counter] << ", ";
}
cout << "}" << endl;
cout << "{";
for (int counter = 0; counter < i; counter++)
{
cout << sales[counter] << ", ";
}
cout << "}" << endl;
return 0;
}
To solve this problem, I need to have the same commands and keywords I still have, and it can't be any advanced or weird syntax. For some reason, my input from the:
cin >> tempSales
Is not functioning. Here are the results:
{mild, medium, sweet, hot, zesty, }
{0,0,0,0,0, }
Whereas I just want to see 1, 2, 3, 4, and 5 for the second array. Why is it only print 0 and not reading my input? Please help!
Like rranjik stated, you shouldn't need a 2D array if you're only listing the number of sales, which it appears you're doing from what you provided, is this not the case?
Is it necessary for you to use the bitshift operator >> for your assignment? For a simple integer assignment it isn't really necessary, and you could do:
int sales[5] = {0}; Change the array to a simple array instead of 2D.
sales[counter] = tempSales; Use standard assignment for the integer on line 19
cout << sales[counter] << ", "; Change your output accordingly.
Hope this helps!
I don't think you need a 2D array for sales. Try cout << sales[counter][5] << ", "; or change int sales[5][6] = {0}; to int sales[5] = {0};. As Luke mentioned, use standard assignment sales[counter] = tempSales;.
I am messing around with dynamic arrays for a user defined amount of inputs for an and gate.
The issue I am running into is that I don't know how many inputs the user is going to test and I need to be able to have an if-else statement that tests each input.
#include <iostream>
#include <iomanip>
#include <string>
using namespace std;
class logic_gate {
public:
int x = 0;
};
int main() {
int userInput = 0;
cout << "How many inputs do you want on your and gate?: ";
cin >> userInput;
cout << endl;
logic_gate *and_gate = new logic_gate[userInput];
cout << endl << "Please enter the values of each bit below . . ." << endl <<
endl;
int userTest1 = 0;
for (int i = 0; i < userInput; i++) {
cout << "#" << i + 1 << ": ";
cin >> userTest1;
and_gate[i].x = userTest1;
}
return 0;
}
Here is the code that I am currently trying to find a solution for.
To implement an AND gate with n inputs you can simply do:
int output = 1;
for (int i = 0; i < n; ++i)
{
if (!and_gate [i])
{
output = 0;
break;
}
}
// ...
Use Vector data structure, you don't need to tell its size while declaring, unlike array, and it can grow automatically.
To read input till it's arriving, put cin inside while loop condition. I used getline to read whole line and work with it, so that whenever user presses enter button at empty line, program will think that no more input is coming anymore, and will start calculating 'And' of inputs.
//don't forget to import vector
#include <iostream>
#include <vector>
#include <string>
using namespace std;
class logic_gate {
public:
int x = 0;
logic_gate(){ //default constructor
}
logic_gate(int k){ //another constructor needed
x = k;
}
};
int main(){
cout << endl << "Please enter the values of each bit below . . ." << endl;
vector<logic_gate> and_gate; //no need to tell size while declaration
string b;
while(getline(cin, b)){ //read whole line from standard input
if (b == "\0") //input is NULL
break;
and_gate.push_back(logic_gate(stoi(b))); //to convert string to integer
}
if (!and_gate.empty()){
int output = and_gate[0].x;
for (int i = 1; i < and_gate.size(); i++){
output = output & and_gate[i].x;
}
cout << "And of inputs is: " << output << endl;
}
else{
cout << "No input was given!\n";
}
return 0;
}
Feel free to ask if some doubts linger
I figured out what I wanted to do. Thanks to everyone who helped and especially Paul Sanders. Below is my final code.
#include <iostream>
using namespace std;
class logic_gate {
public:
int x = 0;
};
int main() {
int userInput;
int output = 1;
cout << "How many inputs do you want on your and gate?: ";
cin >> userInput;
cout << endl;
logic_gate *and_gate = new logic_gate[userInput];
cout << endl << "Please enter the values of each bit below . . ." << endl <<
endl;
int userTest1;
for (int i = 0; i < userInput; i++) {
cout << "#" << i + 1 << ": ";
cin >> userTest1;
and_gate[i].x = userTest1;
}
if (userInput == 1) {
output = userTest1;
cout << "The test of " << userTest1 << " is " << output << endl << endl;
}
else if (userInput > 1) {
for (int i = 0; i < userInput; i++) {
if (!and_gate[i].x)
{
output = 0;
break;
}
}
cout << "The test of ";
for (int i = 0; i < userInput; i++) {
cout << and_gate[i].x;
}
cout << " is " << output << endl << endl;
}
return 0;
}
FizzBuzz program. The user enters numbers separated by a comma. The program reads input and lets the computer know if divisible by 3, 5 or both. When the user enters 15,5,30, the program will only output the first number, 15 and stops there. What am I doing wrong?
void processVector(vector<int> intVector)
{
bool loop;
for (int i = 0; i < intVector.size(); i++)
{
loop = true;
}
}
int main()
{
cout << "Welcome to the FizzBuzz program!" << endl;
cout << "This program will check if the number you enter is divisable by
3, 5, or both." << endl;
cout << "Please enter an array of numbers separated by a comma like so,
5,10,15" << endl;
cin >> userArray;
vector<int> loadVector(string inputString);
istringstream iss(userArray);
vector <int> v;
int i;
while (iss >> i);
{
v.push_back(i);
if (iss.peek() == ',')
iss.ignore();
if (i % 15 == 0)
{
cout << "Number " << i << " - FizzBuzz!" << endl;
}
else if (i % 3 == 0)
{
cout << "Number " << i << " Fizz!" << endl;
}
else if (i % 5 == 0)
{
cout << "Number " << i << " Buzz!" << endl;
}
else
{
cout << "Number entered is not divisable by 3 or 5." << endl;
}
}
system("pause");
}
Here is how I would approach the problem:
#include <iostream>
#include <string>
#include <vector>
int main() {
std::cout << "!!!Hello World!!!" << std::endl; // prints !!!Hello World!!!
std::cout << "Please enter your numbers seperated by a comma (5, 3, 5, 98, 278, 42): ";
std::string userString;
std::getline(std::cin, userString);
std::vector<int> numberV;
size_t j = 0; // beginning of number
for(size_t i = 0; i < userString.size(); i++){
if((userString[i] == ',') || (i == userString.size() -1)){ // could also use strncmp
numberV.push_back(std::stoi(userString.substr(j, i))); // stoi stands for string to int, and .substr(start, end) creates a new string at the start location and ending at the end location
j = i + 1;
}
}
for(size_t n = 0; n < numberV.size(); n++){
std::cout << numberV[n] << std::endl;
}
return(0);
}
This should give you a method to solve the problem (without handling the fizzbuzz part of your program) that I personally find simpler.
The basic form for using functions is:
<return type> <function_name(<inputs)>{
stuff
};
So, a basic function that takes a string and returns a vector (what you are wanting) would be:
std::vector myStringToVector(std::string inputString){
std::vector V;
// your code (see the prior example for one method of doing this)
return(V);
};
It also looks like they want a separate function for outputting your vector values, this could look something like:
void myVectorPrint(std::vector inputVector){
// your code (see prior example for a method of printing out a vector)
};
Thank you #Aaron for the help. Here is the finished code and it works great!
I had to take a little more time researching a few things and trying to understand which order and what to put where in terms of the functions and how to call them. I appreciate all the help as I said I am a noob.
#include "stdafx.h"
#include <iostream>
#include<sstream>
#include<string>
#include<vector>
using namespace std;
vector<int> loadVector(string inputString)
{
stringstream ss(inputString);
vector <int> numberV;
int n;
size_t j = 0; // beginning of number
for (size_t n = 0; n < inputString.size(); n++)
{
if ((inputString[n] == ',') || (n == inputString.size() - 1))
{
numberV.push_back(std::stoi(inputString.substr(j, n)));
j = n + 1;
}
}
return numberV;
}
void processVector(vector<int> intVector)
{
for (int i = 0; i < intVector.size(); i++)
{
int n = intVector.at(i);
if (n % 15 == 0)
{
cout << "Number " << n << " - FizzBuzz!" << endl;
}
else if (n % 3 == 0)
{
cout << "Number " << n << " Fizz!" << endl;
}
else if (n % 5 == 0)
{
cout << "Number " << n << " Buzz!" << endl;
}
else
{
cout << "Number entered is not divisable by 3 or 5." << endl;
}
}
}
int main()
{
cout << "Welcome to the FizzBuzz program." << endl
<< "Please enter an array of numbers separated by comma's (5, 10, 15)"
<< endl;
string inputString;
getline(cin, inputString);
try
{
vector<int> intVector = loadVector(inputString);
processVector(intVector);
}
catch (const exception& e)
{
cout << "Exception caught: '" << e.what() << "'!;" << endl;
}
system("pause");
return 0;
}
My program has to count how many numbers in a range are even and how many of them are odd but I can't seem to figure it out.It kinda works
but when I put numbers in it spouts out nonsense. I'm an extreme nooob when it comes to programing, I think that the problem has to be at line 21 for (i=n; i<=m; i++) { ?
But I'm not sure. I have a programing book but it does not help much,maybe someone can help?
#include <iostream>
using namespace std;
int main()
{
int n;
int m;
int i;
int a;
int b;
cout << "Enter a number that begins interval: ";
cin >> n;
cout << "Enter a number that ends interval: ";
cin >> m;
a=0;
b=0;
for (i=n; i<=m; i++) {
if (i%2 == 0){
a=a+i;
}
else {
b=b+i;
}
}
cout << " unequal numbers: " << a << endl;
cout << " equal numbers: " << b << endl;
Assuming you mean even and odd numbers your problem lies in this code:
for (i=n; i<=m; i++) {
if (i%2 == 0){
a=a+i; // increase number of even numbers by i
}
else {
b=b+i; // increase number of odd numbers by i
}
}
What you might want do to do is add 1 (instead of whatever i is):
for (i = n; i <= m; ++i) {
if (i % 2 == 0)
++a; // increase number of even numbers by one
else
++b; // increase number of odd numbers by one
}
Also I'd suggest using better variable names, for example even and odd instead of a and b and so on. It makes code easier to understand for everybody, even for you.
Just a little more tips. Assigning variables as soon as you declare them is good practice:
int m = 0;
You can declare variable inside of for loop, and in your case there is no need to declare it out of it:
for (int i = n; i <= m; ++i) { ... }
Example how it can change look and clarity of your code:
#include <iostream>
using namespace std;
int main() {
int from = 0,
to = 0,
even = 0,
odd = 0;
cout << "Enter a number that begins interval: ";
cin >> from;
cout << "Enter a number that ends interval: ";
cin >> to;
for (int i = from; i <= to; ++i) {
if (i % 2 == 0)
++even;
else
++odd;
}
cout << " even numbers: " << even << endl;
cout << " odd numbers: " << odd << endl;
return 0; // don't forget this! main is function returning int so it should return something
}
Ok, so as per the new clarification the following should work
#include <iostream>
using namespace std;
int main()
{
int n;
int m;
int i;
int a;
int b;
cout << "Enter a number that begins interval: ";
cin >> n;
cout << "Enter a number that ends interval: ";
cin >> m;
a=0;
b=0;
for (i=n; i<=m; i++) {
if (i%2 == 0){
a++;
}else {
b++;
}
}
cout << " unequal numbers: " << a << endl;
cout << " equal numbers: " << b << endl;
}
So the following changes were done:
The for loop was closed
a = a + i or b = b + i was wrong as you are adding the counter value to the count which should be a++ or b++. Changed that also
The last two lines where you are showing your result was out of the main method, brought them inside the main method
Hope you find this useful.
You don't need to use loop to count even and odd numbers in a range.
#include <iostream>
int main ()
{
int n,m,even,count;
std::cin >> n >> m;
count=m-n+1;
even=(count>>1)+(count&1 && !(n&1));
std::cout << "Even numbers: " << even << std::endl;
std::cout << "Odd numbers: " << count-even << std::endl;
}
#include <iostream>
using namespace std;
int main()
{
int n, i;
cin >> n;
cout << " even : ";
for (i = 1; i <= n * 2; i++)
{
if (i % 2 == 0)
cout << i << " ";
}
cout << " odd : ";
for (i = 1; i <= n * 2; i++)
{
if (i % 2 != 0)
cout << i << " ";
}
return 0;
}
//input n = 5
// output is even : 2 4 6 8 10
// odd : 1 3 5 7 9
#include <iostream>
using namespace std;
int main()
{
int n;
int m;
int i;
int a;
int b;
cout << "Enter a number that begins interval: ";
cin >> n;
cout << "Enter a number that ends interval: ";
cin >> m;
a = 0;
b = 0;
for (i = n; i < = m; i++) {
if (i%2 == 0){
a = a + 1;
} else {
b = b + 1;
}
}
cout << " unequal numbers: " << a << endl;
cout << " equal numbers: " << b << endl;
}
Not sure why you are looping through all the elements (half of them are going to be even and the other half odd). The only case where you have to consider when the interval length is not divisible by two.
using namespace std;
int main()
{
int n;
int m;
int x;
int odds;
int evens;
cout << "Enter a number that begins interval: ";
cin >> n;
cout << "Enter a number that ends interval: ";
cin >> m;
cout << n << " " << m << endl;
x = m - n + 1;
odds = x / 2;
evens = odds;
if (x % 2 != 0) {
if (n % 2 == 0) {
evens++;
} else {
odds++;
}
}
cout << " even numbers: " << evens << endl;
cout << " odd numbers: " << odds << endl;
}
This is a more readable version of #Lassie's answer
So, I'm creating a coin change algorithm that take a Value N and any number of denomination and if it doesn't have a 1, i have to include 1 automatically. I already did this, but there is a flaw now i have 2 matrix and i need to use 1 of them. Is it possible to rewrite S[i] matrix and still increase the size of array.... Also how can i find the max denomination and the second highest and sooo on till the smallest? Should i just sort it out in an highest to lowest to make it easier or is there a simpler way to look for them one after another?
int main()
{
int N,coin;
bool hasOne;
cout << "Enter the value N to produce: " << endl;
cin >> N;
cout << "Enter number of different coins: " << endl;
cin >> coin;
int *S = new int[coin];
cout << "Enter the denominations to use with a space after it" << endl;
cout << "(1 will be added if necessary): " << endl;
for(int i = 0; i < coin; i++)
{
cin >> S[i];
if(S[i] == 1)
{
hasOne = true;
}
cout << S[i] << " ";
}
cout << endl;
if(!hasOne)
{
int *newS = new int[coin];
for(int i = 0; i < coin; i++)
{
newS[i] = S[i];
newS[coin-1] = 1;
cout << newS[i] << " ";
}
cout << endl;
cout << "1 has been included" << endl;
}
//system("PAUSE");
return 0;
}
You could implement it with std::vector, then you only need to use push_back.
std::sort can be used to sort the denominations into descending order, then it's just a matter of checking whether the last is 1 and adding it if it was missing. (There is a lot of error checking missing in this code, for instance, you should probably check that no denomination is >= 0, since you are using signed integers).
#include <iostream> // for std::cout/std::cin
#include <vector> // for std::vector
#include <algorithm> // for std::sort
int main()
{
std::cout << "Enter the value N to produce:\n";
int N;
std::cin >> N;
std::cout << "Enter the number of different denominations:\n";
size_t denomCount;
std::cin >> denomCount;
std::vector<int> denominations(denomCount);
for (size_t i = 0; i < denomCount; ++i) {
std::cout << "Enter denomination #" << (i + 1) << ":\n";
std::cin >> denominations[i];
}
// sort into descending order.
std::sort(denominations.begin(), denominations.end(),
[](int lhs, int rhs) { return lhs > rhs; });
// if the lowest denom isn't 1... add 1.
if (denominations.back() != 1)
denominations.push_back(1);
for (int coin: denominations) {
int numCoins = N / coin;
N %= coin;
if (numCoins > 0)
std::cout << numCoins << " x " << coin << '\n';
}
return 0;
}
Live demo: http://ideone.com/h2SIHs