i'm trying to build a pyramid with numbers between 1 and the inserted number. For example, if i insert 6 to the integer, that the piramid will be as there:
12345654321
234565432
3456543
45654
565
6
I tried using a for loop but i get in any line one or ++ numbers to 6.
This is the code:
#include<stdio.h>
#include <iostream>
#include <conio.h>
int main()
{
int i,j,d;
std::cin >> d;
for(i=1;i<=d;i++)
{
for(j=1;j<=i;j++)
printf("%d",j);
printf("\n");
}
getch();
return 0;
}
How can i solve this problem building a pyramid like the shown.
Since this is homework, I won't paste an algorithm, but here's a few hints:
This 12345654321 can be printed by counting from one to six and then back to one.
This __3456543__ means that for numbers smaller than n, you have to output a _ instead, where n depends on the level you are printing.
Define your loop variables within the loop: for(int i=1;i<=d;i++) ... They are only interesting within the loop, and access outside the loop is usually an error, which is then flagged by the compiler.
There's no need to for the getch(); at the end. When you're in the debugger, you can put a breakpoint on the last line. If you aren't you don't want to have to press a key just to end your program.
If you use std::cout << j and std::cout << '\n' for outputting, you don't need printf() either. (Once you want formatting, many will tell you that printf format strings are easier. I don't believe that, but would accept it, if it weren't that you can crash any application with an ill-formed printf format string, while it's much harder to come up with a way to crash your app using streams.)
There you go:
for(j=i;j<=d;j++)
Also you forgot about formatting and the right side of the pyramid, but I think that's out of scope for this question and you can figure the code yourself :)
Try to have the first line working the way you want.
Repeat it d times, where d is the number entered by the user.
Notice that on line l, numbers < l are replaced by spaces.
Consider this: you have D rows, 1..D. Six rows means your rows are numbered 1 to 6. Now, for each row d:
print d-1 space characters. First row has no spaces, second has one and so on.
print the numbers d..D. So on the first line you print 1..6, on the second you print 2..6.
print the numbers D-1..d. So on the first line you print 5..1, on the second you print 5..2
print a new line
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.
So, lets say we have a text.txt file with these numbers:
4 5 15 10 20
5 5 15 10 20 25
In the above example, the first numbers in the row describe how many numbers are in that row. The rest of the numbers are the numbers I am interested in (I will be sorting them in a later part of the code, but that is not where my question focuses).
My issue is, how can I best go about taking each row of numbers (ignoring the first number), placing them into a array, and then moving onto the next line and doing the same thing (placing them into an array, that will be later sorted)?
All my google searching points to doing this with strings via getline, and nothing really points to handling it with ints. Hope someone on here can help point me in the right direction.
Below is the basic code I would use to open the file:
#include <iostream>
#include <fstream>
using namespace std;
int main()
{
int a;
ifstream inputfile;
//declare an input file
inputfile.open("text.txt");
while(//not sure best way to do this part)
{
//guessing I can use a for loop and place numbers in array
//based on first number in the row of numbers
}
return 0;
}
The most obvious way would probably be to read a line with std::getline, then put the string into a stringstream, and read numbers from there (ignoring the first, obviously).
I suggest the following approach:
std::string line;
// Keep reading lines of text from the file.
// Break out of the loop when there are no more lines in the file.
while (getline(inputfile, line)){
// Construct a istingstream from the line of text.
std::istringstream ss(line);
// Read the numbers from the istingstream.
// Process the numbers as you please.
}
If I have a text file which only has numbers inside. Such as:
1
2
3
4
5
6
How can I, for example, multiply each number by two and get the result of each individual operation as output to my screen? Would I have to set each number to a variable?
The best solution to your problem is to follow these steps:
Open a file, and create an int variable ( for instance a)
In Do...while loop take number from a file like this
Filehandler >> a;
multiply the a like this
a = a * 2;
or do whatever you want with it.
Print an value
cout << a;
till you get EOF
Of course there is another possibility like storing each value in array and then multiplying it. It depends on you what you will choose.
Do you know for Loops for example.
you can read the the numbers and store them in an array and use a for loop to multiply or div or whatever and print that .
In my program, there will be two lines of output: numbers in the first line are integers less than 1000; numbers in the second line are floats between 0 and 10. The format of the output I want is like:
right format
and here is how I try (all outputs are the last four lines):
case 1 - 5
output without format; in the picture we can see the numbers that should be printed
line 1 without format and line 2 with format
numbers in line 1; only the first one is right, and the number that is bigger than 100 is in scientific notation
both lines with format
just like case 2, and <<setw(3)<<setfill('0') seems only works for one time to change 43 to 043, and then it won't work any more
With noshowpoint added. If i print *(p + 2)alone, it always prints 118with or without format. But if I print *(p + 2) in the for loop, the output is still 1.2e+002 except for the first one. But if I print the number 118 directly, the output is right.
So, which part of my code is wrong? And what is the right way to output the answer in right format?
Indeed, not all IO Formatting Manipulators are persistent. Some of them, like std::setw only take effect on the next operation, then they expire. There's nothing "wrong" with your code; you just need to use the manipulator again. Refer to your favourite C++ standard library documentation to check the persistence of each manipulator that you use.
Well, for the third, you set showpoint but never unset it with the noshowpoint, hence the next lines issue.
#include <iostream>
#include <iomanip>
using namespace std;
int main() {
double x[4]= {43.0,118.0,9.0,5.0};
double y[4] = {9.5, 9.0, 8.5, 8.0};
// your code goes here
for (int i=0; i< 4; i++)
{
cout<<noshowpoint<<setprecision(3)<<setw(3)<<setfill('0')<<(x[i]) <<" ";
cout<<showpoint<<setprecision(2)<<y[i] << endl;
}
return 0;
}
gives the desired output.
Just pay attention that you set the showpoint and setprecision manipulators for stream, not for the single output.
In my c++ class, we got assigned pairs. Normally I can come up with an effective algorithm quite easily, this time I cannot figure out how to do this to save my life.
What I am looking for is someone to explain an algorithm (or just give me tips on what would work) in order to get this done. I'm still at the planning stage and want to get this code done on my own in order to learn. I just need a little help to get there.
We have to create histograms based on a 4 or 5 integer input. It is supposed to look something like this:
Calling histo(5, 4, 6, 2) should produce output that appears like:
*
* *
* * *
* * *
* * * *
* * * *
-------
A B C D
The formatting to this is just killing me. What makes it worse is that we cannot use any type of arrays or "advanced" sorting systems using other libraries.
At first I thought I could arrange the values from highest to lowest order. But then I realized I did not know how to do this without using the sort function and I was not sure how to go on from there.
Kudos for anyone who could help me get started on this assignment. :)
Try something along the lines of this:
Determine the largest number in the histogram
Using a loop like this to construct the histogram:
for(int i = largest; i >= 1; i--)
Inside the body of the loop, do steps 3 to 5 inclusive
If i <= value_of_column_a then print a *, otherwise print a space
Repeat step 3 for each column (or write a loop...)
Print a newline character
Print the horizontal line using -
Print the column labels
Maybe i'm mistaken on your q, but if you know how many items are in each column, it should be pretty easy to print them like your example:
Step 1: Find the Max of the numbers, store in variable, assign to column.
Step 2: Print spaces until you get to column with the max. Print star. Print remaining stars / spaces. Add a \n character.
Step 3: Find next max. Print stars in columns where the max is >= the max, otherwise print a space. Add newline. at end.
Step 4: Repeat step 3 (until stop condition below)
when you've printed the # of stars equal to the largest max, you've printed all of them.
Step 5: add the -------- line, and a \n
Step 6: add row headers and a \n
If I understood the problem correctly I think the problem can be solved like this:
a= <array of the numbers entered>
T=<number of numbers entered> = length(a) //This variable is used to
//determine if we have finished
//and it will change its value
Alph={A,B,C,D,E,F,G,..., Z} //A constant array containing the alphabet
//We will use it to print the bottom row
for (i=1 to T) {print Alph[i]+" "}; //Prints the letters (plus space),
//one for each number entered
for (i=1 to T) {print "--"}; //Prints the two dashes per letter above
//the letters, one for each
while (T!=0) do {
for (i=1 to N) do {
if (a[i]>0) {print "*"; a[i]--;} else {print " "; T--;};
};
if (T!=0) {T=N};
}
What this does is, for each non-zero entered number, it will print a * and then decrease the number entered. When one of the numbers becomes zero it stops putting *s for its column. When all numbers have become zero (notice that this will occur when the value of T comes out of the for as zero. This is what the variable T is for) then it stops.
I think the problem wasn't really about histograms. Notice it also doesn't require sorting or even knowing the