I have been doing exercises on some online judges, and I encounter this question with this default answer.
Question description:
Finally, Hansbug has finally reached the moment to do the last math problem, and there are a bunch of messy addition and subtraction equations in front of him. Obviously success is at hand. But his brain cells have been exhausted, so this important task is left to you.
Input format:
One line, containing a string of addition and subtraction polynomials (the range of each item is 0-32767).
Output format:
An integer, which is the result of the calculation (guarantee that the result of the calculation will not exceed the range of the long integer).
Input and output sample:
Enter #1:
1+2-3
Output #1:
0
And the default answer is :
#include<bits/stdc++.h>
using namespace std;
int ans;
int c;
int main() {
while (cin >> c)
ans += c;
cout << ans;
return 0;
}
How is this even possible!?
All right, let's try it, let's put there some expression with addition and subtraction of ints only, after that press Return, then ctrl+D (end of input):
$ ./a.out
111-222+1
-110$
The loop while (cin >> c) will parse integers including the sign one by one using iostream capabilities until an end of input (you also have to terminate the last number by pressing Return, effectively putting the newline there and triggering the last cin >> c).
Related
Ok so i have been learning binary search. My teacher has given me a problem on codeforces and it always fails on test 2. Here is the problem:
In this problem jury has some number x, and you have to guess it. The number x is always an integer from 1 and to n, where n
is given to you at the beginning.
You can make queries to the testing system. Each query is a single integer from 1
to n
. Flush output stream after printing each query. There are two different responses the testing program can provide:
the string "<" (without quotes), if the jury's number is less than the integer in your query;
the string ">=" (without quotes), if the jury's number is greater or equal to the integer in your query.
When your program guessed the number x
, print string "! x", where x
is the answer, and terminate your program normally immediately after flushing the output stream.
Your program is allowed to make no more than 25 queries (not including printing the answer) to the testing system.
Input
Use standard input to read the responses to the queries.
The first line contains an integer n
(1≤n≤pow(10,6) — maximum possible jury's number.
Following lines will contain responses to your queries — strings "<" or ">=". The i
-th line is a response to your i-th query. When your program will guess the number print "! x", where x
is the answer and terminate your program.
The testing system will allow you to read the response on the query only after your program print the query for the system and perform flush operation.
Output
To make the queries your program must use standard output.
Your program must print the queries — integer numbers xi
(1≤xi≤n), one query per line (do not forget "end of line" after each xi
). After printing each line your program must perform operation flush.
And here is my code:
#include <iostream>
using namespace std;
int main()
{
int n;
string s;
int k=0;
cin>>n;
int min=1,max=n;
int a;
while(k==0)
{
if(max==min+1)
{
cout<<"! "<<min;
k=1;
break;
}
a=(min+max)/2;
cout<<a<<endl;
cin>>s;
if(s==">=")
min=a;
else
max=a;
}
}
I dont know what test 2 is, but i would be happy to hear ideas as to where my programs is wrong. My guess is its something with the number of guesses. Thanks in advance!
I have tried variations of the loop written above, but they all give the same result.
I managed to get this program to work. If the user types an unfixed amount of integers, the program will calculate the average value of it. But I need to end it with <Ctrl-D> in my terminal (end of file) in order for it to work. Why can I not just press enter for it to work?
I also believe that I've used an unnecessary amount of variables. Can it be narrowed down to maybe 2 variables?
#include <iomanip>
#include <iostream>
using namespace std;
int main ()
{
int digit {};
int res {};
int counter {};
cout << "Type in integers: ";
while (cin >> digit)
{
counter ++;
res += digit;
}
cout << "The mean was " << setw(1) << setprecision(1) << fixed << static_cast<double>(res) / static_cast<double>(counter) << endl;
return 0;
}
Why can I not just press enter for it to work?
Because that's not how the overloaded >> formatted extraction operator works. This operator skips over an unlimited amount of whitespace characters, including newline characters, until it reads the integer. It's simply how it works: it will read newlines and spaces, after newlines, and spaces, until it sees a digit. That's its mission in life: read and skip over spaces and newlines until it reads at least one digit. It never gets tired of reading newlines and spaces, and will keep going as long that's the case.
To handle input in the fashion you describe requires a completely different approach: using std::getline to read a single line of input into a std::string, up until the next newline character. Then, once that's done, you can check if the std::string is empty, which means that no input was entered, and then terminate; otherwise take the input in std::string and convert it to an int value (using std::stoi, std::from_chars, or a std::istringstream -- take your pick), and then proceed with the existing algorithm.
Can it be narrowed down to maybe 2 variables?
How do you expect to do that? Hard, immutable logic dictates that you must keep track of at least two discrete values: the total sum and the number of values read. But then you just ran out of variables. You have no more variables to use for storing the next read value (if there is one), using whatever approach you chose to use. So, you can't do it. Rules of logic require the use of at least three variables, possibly more depending on how fancy and robust you want your input validation to work.
I am trying to read the last integer from an input such as-
100 121 13 ... 7 11 81
I'm only interested in the last integer and hence want to ignore all
previous integers.
I thought of using cin.ignore but that won't work here due to
unknown integers (100 is of 3 digits, while 13 is of 2 digits & so on)
I can input integer by integer using a loop and do nothing with them. Is there a better way?
It all depends on the use case that you have.
Reading a none specified number of integers from std::cin is not as easy at it may seem. Because, in contrast to reading from a file, you will not have an EOF condition. If you would read from a file stream, then it would be very simple.
int value{};
while (fileStream >> value)
;
If you are using std::cin you could try pressing CTRL-D or CTRL-Z or whatever works on your terminal to produce an EOF (End Of File) condition. But usually the approach is to use std::getline to read a complete line until the user presses enter, then put this line into a std::istringstream and extract from there.
Insofar, one answer given below is not that good.
So, next solution:
std::string line{};
std::getline(std::cin, line);
std::istringstream iss{line};
int value{};
while (iss >> value)
;
You were asking
Is there a better way?
That also depends a little. If you are just reading some integers, then please go with above approach. If you would have many many values, then you would maybe waste time by unnecessarily converting many substrings to integers and loose time.
Then, it would be better, to first read the complete string, then use rfind to find the last space in the string and use std::stoi to convert the last substring to an integer.
Caveat: In this case you must be sure (or check with more lines of code) that there are no white space at the end and the last substring is really a number. That is a lot of string/character fiddling, which can most probably avoided.
So, I would recommend the getline-stringstream approach.
You can try this simple solution for dynamically ignoring rest of the values except the last given in this problem as shown:
int count = 0;
int values, lastValue; // lastValue used for future use
std::cout << "Enter your input: ";
while (std::cin >> values) {
lastValue = values; // must be used, otherwise values = 0 when loop ends
count++;
}
std::cout << lastValue; // prints
Note: A character must be required to stop the while(), hence it's better put a . at last.
Output example
Enter your input: 3 2 4 5 6 7.
7
Try this:
for( int i=0; i<nums_to_ignore; i++) {
int ignored;
std::cin >> ignored;
}
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)
Input
The input begins with two positive integers n k (n, k<=10^7). The next n lines of input contain one positive integer ti, not greater than 10^9, each.
Output
Write a single integer to output, denoting how many integers ti are divisible by k.
Example
Input:
7 3
1
51
966369
7
9
999996
11
Output:
4
My Code:
#include <iostream>
using namespace std;
int main()
{
long long n,k, i;
cin>>n;
cin>>k;
int count=0;
for(i=0;i<n;i++)
{
int z;
cin>>z;
if(z%k == 0) count++;
}
cout<<count;
return 0;
}
Now this code produces the correct output. However, its not being accepted by CodeChef(http://www.codechef.com/problems/INTEST) for the following reason: Time Limit Exceeded. How can this be further optimized?
As said by caleb the problem is labeled "Enormous Input Test" so it requires you to use some better/faster I/O methods
just replacing cout with printf and cin with scanf will give you an AC but to improve your execution time you need to use some faster IO method for example reading character by character using getchar_unlocked() will give you a better execution time
so you can read the values by using a function like this , for a better execution time.
inline int read(){
char c=getchar_unlocked();
int n=0;
while(!(c>='0' && c<='9'))
c=getchar_unlocked();
while(c>='0' && c<='9'){
n=n*10 + (c-'0');
c=getchar_unlocked();
}
return n;
}
The linked problem contains the following description:
The purpose of this problem is to verify whether the method you are
using to read input data is sufficiently fast to handle problems
branded with the enormous Input/Output warning. You are expected to be
able to process at least 2.5MB of input data per second at runtime.
Considering that, reading values from input a few bytes at a time using iostreams isn't going to cut it. I googled around a bit and found a drop-in replacement for cin and cout described on CodeChef. Some other approaches you could try include using a memory-mapped file and using stdio.
It might also help to look for ways to optimize the calculation. For example, if ti < k, then you know that k is not a factor of ti. Depending on the magnitude of k and the distribution of ti values, that observation alone could save a lot of time.
Remember: the fact that your code is short doesn't mean that it's fast.