How can we print given matrix spirally from any point and specified direction?
for example
if given matrix is
21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
let current position be pointing to 1 and required direction is clockwise
correct output should be 1 2 3 4 5 6 .... upto 25.
I am looking for logic and code in C or C++
Related
>list1=[1,2,3,4]
>list2=[5,6,7,8]
>list3=[9,10,11,12]
>list4=[13,14,15,16]
>list5=[17,18,19,20]
>lists=[list1,list2,list3,list4,list5
I want to print the following code so that it outputs this way:
4 8 12 16 20
3 7 11 15 19
2 6 10 14 18
sorry didn't knew it ignored new lines:
1 5 9 13 17
Thanks in advance (new to python)
One way you could achieve this is to zip up the reversed lists and simply print all the elements out.
list1=[1,2,3,4]
list2=[5,6,7,8]
list3=[9,10,11,12]
list4=[13,14,15,16]
list5=[17,18,19,20]
for l1, l2, l3, l4, l5 in zip(reversed(list1), reversed(list2), reversed(list3), reversed(list4), reversed(list5)):
print(l1, l2, l3, l4, l5, end=' ')
output
4 8 12 16 20 3 7 11 15 19 2 6 10 14 18 1 5 9 13 17
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I have a for loop and a variable C.
The loop begins at 0 and is expected to run C times but actually it runs C-1 times.
Here is my code :
vector<int> v(1000) //Allocated globally
int M, S, C;
cin>>M>>S>>C;
//cout<<M<<" "<<S<<" "<<C;
int fi=0, la=0;
for(int i=0; i<C; i++)
{
int f;
cin>>f;
if(i==0l){ fi = f;}
v[f] = f;
cout<<i<<" "<<f<<" "<<v[f]<<endl;
if(i==C-1){ la = f;}
}
This is my test case -
3 27 16
2
3
5
6
8
9
10
13
14
15
16
19
20
21
22
27
Output by Xcode :
0 2 2
1 3 3
2 5 5
3 6 6
4 8 8
5 9 9
6 10 10
7 13 13
8 14 14
9 15 15
10 16 16
11 19 19
12 20 20
13 21 21
14 22 22
I use Xcode on Mac if it makes a difference.
The variables fi and la are to find the first and the last element of the list.
I want to know what is wrong in my code for the for loop and why is it not iterating C times.
Thanks
Your loop is iterating C times. This is the classic Zero-Based Numbering issue.
Let me explain using your example where C is 16 and a numbered list:
2
3
5
6
8
9
10
13
14
15
16
19
20
21
22
27
So you see your lopp did iterate 16 times. To go from 0 to 16 would have actually been iterating one more time, so 17 times.
I want to compare two csv files looking like below.
Here I want to find out unmatched signals.
I need some help in python. Please help me.
File 1
2
USER Name
7/31/2015 0:00
<XXXXXXX>
1 Signal_1 10
2 Signal_2 1 2 3 4 5
3 Signal_3 X 5 10 15 20 25 Y 6 11 16 21 26
1 Signal_4 20
1 Signal_5 30
2 Signal_6 6 7 8 9 10 11 12 13
2 Signal_7 55 1.05 1.6 14.1
3 Signal_8 X 30 40 50 60 40 Y 14 15 26 14 26
2 Signal_9 1 1 2 3 2
1 Signal_10 40
File 2
2
USER Name
7/31/2015 0:00
<XXXXXXX>
3 Signal_3 X 20 10 15 17 25 Y 6 11 16 21 26
1 Signal_5 5
2 Signal_7 55 1.05 1.6 14.1
1 Signal_1 10
3 Signal_8 X 30 40 50 60 40 Y 14 15 26 14 26
1 Signal_10 14
2 Signal_9 1 1 2 3 2
2 Signal_6 6 7 8 59 10 15 12 13
1 Signal_4 20
2 Signal_2 1 2 3 4 5
Result:
File
3 Signal_3 X 5 10 15 20 25 Y 6 11 16 21 26
1 Signal_5 30
1 Signal_10 40
2 Signal_6 6 7 8 9 10 11 12 13
File 2
3 Signal_3 X 20 10 15 17 25 Y 6 11 16 21 26
1 Signal_5 5
1 Signal_10 14
2 Signal_9 1 1 2 3 2
If you want to check for fairly exact comparisons, you can use sets quite easily:
def sigset(fname):
with open(fname, 'rb') as f:
data = set(' '.join(line.split()) for line in f
if 'Signal' in line)
return data
s1 = sigset('sig1.txt')
s2 = sigset('sig2.txt')
print('File 1')
for line in sorted(s1 - s2):
print(line)
print('')
print('File 2')
for line in sorted(s2 - s1):
print(line)
with open('Sample1.csv', 'r') as t1, open('Sample2.csv', 'r') as t2:
fileone = t1.readlines()
filetwo = t2.readlines()
print fileone
print filetwo
with open('update.csv', 'w') as outFile:
for line in filetwo:
if line not in fileone:
outFile.write(line)
for line in fileone:
if line not in filetwo:
outFile.write(line)
I've made a program to create the pascal's triangle. the program takes number of rows as input and displays the triangle on the console. I've used the setw() function to set the distance between numbers. it's of for unit single digits but when the numbers get greater than 10,the width is not being adjusted properly,right now I've :
if(P<10){
std::cout << P ;
std::cout <<std::setw(2);
}
if(P>=10){
std::cout<<std::setw(3) << P ;
std::cout<<std::setw(2);
}
here's the ouput from the console:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84126126 84 36 9 110
I want it to appear like a proper triangle,Could someone help me out please???
If you read e.g. this reference of std::setw you will see
The width property of the stream will be reset to zero (meaning "unspecified") if any of the following functions are called
And then goes on to list basically all output operators.
This means that when you do
std::cout <<std::setw(2);
the width will only be set for the next output operation. If you do any kind of output after that the width will be reset to zero.
I'm trying to search a big project for all examples of where I've declared an array with [48] as the size or any multiples of 48.
Can I use a regular expression function to find matches of 48 * n?
Thanks.
Here you go (In PHP's PCRE syntax):
^(0*|(1(01*?0)*?1|0)+?0{4})$
Usage:
preg_match('/^(0*|(1(01*?0)*?1|0)+?0{4})$/', decbin($number));
Now, why it works:
Well we know that 48 is really just 3 * 16. And 16 is just 2*2*2*2. So, any number divisible by 2^4 will have the 4 most bits in its binary representation 0. So by ending the regexp with 0{4}$ is equivalent to saying that the number is divisible by 2^4 (or 16). So then, the bits to the left need to be divisible by 3. So using the regexp from this answer, we can tell if they are divisible by 3. So if the whole regexp matches, the number is divisible by both 3 and 16, and hence 48...
QED...
(Note, the leading 0| case handles the failed match when $number is 0). I've tested this on all numbers from 0 to 48^5, and it correctly matches each time...
A generalization of your question is asking whether x is a string representing a multiple of n in base b. This is the same thing as asking whether the remainder of x divided by n is 0. You can easily create a DFA to compute this.
Create a DFA with n states, numbered from 0 to n - 1. State 0 is both the initial state and the sole accepting state. Each state will have b outgoing transitions, one for each symbol in the alphabet (since base-b gives you b digits to work with).
Each state represents the remainder of the portion of x we've seen so far, divided by n. This is why we have n of them (dividing a number by n yields a remainder in the range 0 to n - 1), and also why state 0 is the accepting state.
Since the digits of x are processed from left to right, if we have a number y from the first few digits of x and read the digit d, we get the new value of y from yb + d. But more importantly, the remainder r changes to (rb + d) mod n. So we now know how to connect the transition arcs and complete the DFA.
You can do this for any n and b. Here, for example, is one that accepts multiples of 18 in base-10 (states on the rows, inputs on the columns):
| 0 1 2 3 4 5 6 7 8 9
---+-------------------------------
→0 | 0 1 2 3 4 5 6 7 8 9 ←accept
1 | 10 11 12 13 14 15 16 17 0 1
2 | 2 3 4 5 6 7 8 9 10 11
3 | 12 13 14 15 16 17 0 1 2 3
4 | 4 5 6 7 8 9 10 11 12 13
5 | 14 15 16 17 0 1 2 3 4 5
6 | 6 7 8 9 10 11 12 13 14 15
7 | 16 17 0 1 2 3 4 5 6 7
8 | 8 9 10 11 12 13 14 15 16 17
9 | 0 1 2 3 4 5 6 7 8 9
10 | 10 11 12 13 14 15 16 17 0 1
11 | 2 3 4 5 6 7 8 9 10 11
12 | 12 13 14 15 16 17 0 1 2 3
13 | 4 5 6 7 8 9 10 11 12 13
14 | 14 15 16 17 0 1 2 3 4 5
15 | 6 7 8 9 10 11 12 13 14 15
16 | 16 17 0 1 2 3 4 5 6 7
17 | 8 9 10 11 12 13 14 15 16 17
These get really tedious as n and b get larger, but you can obviously write a program to generate them for you no problem.
1|48|2304|110592|5308416
You are unlikely to have declared an array of size 48^5 or larger.
No, regular expressions can't calculate multiples (except in the unary number system: decimal 4 = unary 1111; decimal 8 = unary 11111111, so the regex ^(1111)+$ matches multiples of 4).
import re
# For real example,
# construction of a chain with integers multiples of 48
# and integers not multiple of 48.
from random import *
w = [ 48*randint( 1,10) for j in xrange(10) ]
w.extend( 48*randint(11,20) for j in xrange(10) )
w.extend( 48*randint(21,70) for j in xrange(10) )
a = [ el if el%48!=0 else el+1 for el in sample(xrange(1000),40) ]
w.extend(a)
shuffle(w)
texte = [ ''.join(sample(' abcdefghijklmonopqrstuvwxyz',randint(1,7))) for i in xrange(40) ]
X = ''.join(texte[i]+str(w[i]) for i in xrange(40))
# Searching the multiples of 48 in the chain X
def mult48(match):
g1 = match.group()
if int(g1)%48==0:
return ( g1, X[0:match.end()] )
else:
return ( g1, 'not multiple')
for match in re.finditer('\d+',X):
print '%s %s\n' % mult48(match)
Any multiple is difficult, but here's a (python-style) regexp that matches the first 200 multiples of 48.
0$|1(?:0(?:08$|56$)|1(?:04$|52$)|2(?:00$|48$|96$)|3(?:44$|92$)|4(?:4(?:$|0$)|88$\
)|5(?:36$|84$)|6(?:32$|80$)|7(?:28$|76$)|8(?:24$|72$)|9(?:2(?:$|0$)|68$))|2(?:0(\
?:16$|64$)|1(?:12$|60$)|2(?:08$|56$)|3(?:04$|52$)|4(?:0(?:$|0$)|48$|96$)|5(?:44$\
|92$)|6(?:40$|88$)|7(?:36$|84$)|8(?:32$|8(?:$|0$))|9(?:28$|76$))|3(?:0(?:24$|72$\
)|1(?:20$|68$)|2(?:16$|64$)|3(?:12$|6(?:$|0$))|4(?:08$|56$)|5(?:04$|52$)|6(?:00$\
|48$|96$)|7(?:44$|92$)|8(?:4(?:$|0$)|88$)|9(?:36$|84$))|4(?:0(?:32$|80$)|1(?:28$\
|76$)|2(?:24$|72$)|3(?:2(?:$|0$)|68$)|4(?:16$|64$)|5(?:12$|60$)|6(?:08$|56$)|7(?\
:04$|52$)|8(?:$|0(?:$|0$)|48$|96$)|9(?:44$|92$))|5(?:0(?:40$|88$)|1(?:36$|84$)|2\
(?:32$|8(?:$|0$))|3(?:28$|76$)|4(?:24$|72$)|5(?:20$|68$)|6(?:16$|64$)|7(?:12$|6(\
?:$|0$))|8(?:08$|56$)|9(?:04$|52$))|6(?:0(?:00$|48$|96$)|1(?:44$|92$)|2(?:4(?:$|\
0$)|88$)|3(?:36$|84$)|4(?:32$|80$)|5(?:28$|76$)|6(?:24$|72$)|7(?:2(?:$|0$)|68$)|\
8(?:16$|64$)|9(?:12$|60$))|7(?:0(?:08$|56$)|1(?:04$|52$)|2(?:0(?:$|0$)|48$|96$)|\
3(?:44$|92$)|4(?:40$|88$)|5(?:36$|84$)|6(?:32$|8(?:$|0$))|7(?:28$|76$)|8(?:24$|7\
2$)|9(?:20$|68$))|8(?:0(?:16$|64$)|1(?:12$|6(?:$|0$))|2(?:08$|56$)|3(?:04$|52$)|\
4(?:00$|48$|96$)|5(?:44$|92$)|6(?:4(?:$|0$)|88$)|7(?:36$|84$)|8(?:32$|80$)|9(?:2\
8$|76$))|9(?:0(?:24$|72$)|1(?:2(?:$|0$)|68$)|2(?:16$|64$)|3(?:12$|60$)|4(?:08$|5\
6$)|5(?:04$|52$)|6(?:$|0$))