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.
Related
I have the following data in Stata:
clear
* Input data
input grade id exit time
1 1 . 10
2 1 . 20
3 1 2 30
4 1 0 40
5 1 . 50
1 2 0 10
2 2 0 20
3 2 0 30
4 2 0 40
5 2 0 50
1 3 1 10
2 3 1 20
3 3 0 30
4 3 . 40
5 3 . 50
1 4 . 10
2 4 . 20
3 4 . 30
4 4 . 40
5 4 . 50
1 5 1 10
2 5 2 20
3 5 1 30
4 5 1 40
5 5 1 50
end
The objective is to take the first row foreach id when a event occurs and if no event occur then take the last report foreach id. Here is a example for the data I hope to attain
* Input data
input grade id exit time
3 1 2 30
5 2 0 50
1 3 1 10
5 4 . 50
1 5 1 10
end
The definition of an event appears to be that exit is not zero or missing. If so, then all you need to do is tweak the code in my previous answer:
bysort id (time): egen when_first_e = min(cond(exit > 0 & exit < ., time, .))
by id: gen tokeep = cond(when_first_e == ., time == time[_N], time == when_first_e)
Previous thread was here.
I am writing a program to examine the string STRING to see where it matches SUBSTRING using gawk. One problem I have run into is that the match function only gives the left most match in the string. My current thought is to use gsub to find out how many times the SUBSTRING is present and then use match multiple times using the last substring(STRING,RSTART+1) to find the true start positions of each position, of course with some edits to the code. I am wondering if there is an easier way than this, or a built in function that gives all RSTARTS.
Example:
STRING=DDDADDCDFFDFGSDD
SUBSTRING=D
EDIT:
I looked at the array function for match (thanks for pointing me to more up to date documentation than I had been reading). This still doesn't work, as it allows you to search for multiple things in the same string, but still only gives the left most location of each of these strings.
For example:
$ echo DDDADDCDFFDFGSDD | gawk '{match($0,/D/,a); for (i in a) print i,a[i]}'
0start 1
0length 1
0 D
it works to find the left most of multiple things
echo gDDDADDCDFFDFGSDD | gawk '{match($0,/(D)(A)/,a); for (i in a) print i,a[i]}'
0start 4
0length 2
1start 4
2start 5
2length 1
1length 1
0 DA
1 D
2 A
So we are still finding the left most match (which is what the documentation say it will do)
There isn't a native way to deal with this that i have found, so I wrote this function to do it. This will only work with version of gawk that allow for multidimensional arrays, though making this work with older versions of awk would be simple as well, though parsing afterwards would be more difficult.
The function searches through the string for the regex and populates an array MM. It returns -1 if there was an error, 0 if there were no matches found, else it returns the number of matches found.
function multiMatch(string,subs){
split("",MM,"")
RLENGTH=0
RSTART=0
t=0
s=string
if (length(string) == 0 || length(subs) == 0){
print "Must have string and Regex to look for"
return -1
}
while (1) {
t=RSTART+t
s=substr(string,t+1)
if ( length(s) == 0 ){
break
}
match(s,subs)
if (RLENGTH == -1) {
break
}
found=substr(string,0,length(string)-(length(string)-t-RSTART+1))"-"substr(string,t+RSTART,RLENGTH)"-"substr(string,t+RSTART+RLENGTH);
MM[n]["RSTART"]=RSTART
MM[n]["RLENGTH"]=RLENGTH
MM[n]["STR"]=found
n++
}
return n
}
Example
echo doogggogogggggggooogggogggggooogoooggoooo g*o | awk '
BEGIN{PROCINFO["sorted_in"]="#ind_num_asc"}
{
print "Found "multiMatch($1,$2)" Matches"
for (x in MM) {
print x,MM[x]["RSTART"],MM[x]["RLENGTH"],MM[x]["STR"]
}
}'
OUTPUT
Found 40 Matches
2 1 d-o-ogggogogggggggooogggogggggooogoooggoooo
1 1 1 do-o-gggogogggggggooogggogggggooogoooggoooo
2 1 4 doo-gggo-gogggggggooogggogggggooogoooggoooo
3 1 3 doog-ggo-gogggggggooogggogggggooogoooggoooo
4 1 2 doogg-go-gogggggggooogggogggggooogoooggoooo
5 1 1 dooggg-o-gogggggggooogggogggggooogoooggoooo
6 1 2 doogggo-go-gggggggooogggogggggooogoooggoooo
7 1 1 doogggog-o-gggggggooogggogggggooogoooggoooo
8 1 8 doogggogo-gggggggo-oogggogggggooogoooggoooo
9 1 7 doogggogog-ggggggo-oogggogggggooogoooggoooo
10 1 6 doogggogogg-gggggo-oogggogggggooogoooggoooo
11 1 5 doogggogoggg-ggggo-oogggogggggooogoooggoooo
12 1 4 doogggogogggg-gggo-oogggogggggooogoooggoooo
13 1 3 doogggogoggggg-ggo-oogggogggggooogoooggoooo
14 1 2 doogggogogggggg-go-oogggogggggooogoooggoooo
15 1 1 doogggogoggggggg-o-oogggogggggooogoooggoooo
16 1 1 doogggogogggggggo-o-ogggogggggooogoooggoooo
17 1 1 doogggogogggggggoo-o-gggogggggooogoooggoooo
18 1 4 doogggogogggggggooo-gggo-gggggooogoooggoooo
19 1 3 doogggogogggggggooog-ggo-gggggooogoooggoooo
20 1 2 doogggogogggggggooogg-go-gggggooogoooggoooo
21 1 1 doogggogogggggggoooggg-o-gggggooogoooggoooo
22 1 6 doogggogogggggggooogggo-gggggo-oogoooggoooo
23 1 5 doogggogogggggggooogggog-ggggo-oogoooggoooo
24 1 4 doogggogogggggggooogggogg-gggo-oogoooggoooo
25 1 3 doogggogogggggggooogggoggg-ggo-oogoooggoooo
26 1 2 doogggogogggggggooogggogggg-go-oogoooggoooo
27 1 1 doogggogogggggggooogggoggggg-o-oogoooggoooo
28 1 1 doogggogogggggggooogggogggggo-o-ogoooggoooo
29 1 1 doogggogogggggggooogggogggggoo-o-goooggoooo
30 1 2 doogggogogggggggooogggogggggooo-go-ooggoooo
31 1 1 doogggogogggggggooogggogggggooog-o-ooggoooo
32 1 1 doogggogogggggggooogggogggggooogo-o-oggoooo
33 1 1 doogggogogggggggooogggogggggooogoo-o-ggoooo
34 1 3 doogggogogggggggooogggogggggooogooo-ggo-ooo
35 1 2 doogggogogggggggooogggogggggooogooog-go-ooo
36 1 1 doogggogogggggggooogggogggggooogooogg-o-ooo
37 1 1 doogggogogggggggooogggogggggooogoooggo-o-oo
38 1 1 doogggogogggggggooogggogggggooogoooggoo-o-o
39 1 1 doogggogogggggggooogggogggggooogoooggooo-o-
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++
I'm trying to implement an ACO for 01MKP. My input values are from the OR-Library mknap1.txt. According to my algorithm, first I choose an item randomly. then i calculate the probabilities for all other items on the construction graph. the probability equation depends on pheremon level and the heuristic information.
p[i]=(tau[i]*n[i]/Σ(tau[i]*n[i]).
my pheremon matrix's cells have a constant value at initial (0.2). for this reason when i try to find the next item to go, pheremon matrix is becomes ineffective because of 0.2. so, my probability function determines the next item to go, checking the heuristic information. As you know, the heuristic information equation is
n[i]=profit[i]/Ravg.
(Ravg is the average of the resource constraints). for this reason my prob. functions chooses the item which has biggest profit value. (Lets say at first iteration my algorithm selected an item randomly which has 600 profit. then at the second iteration, chooses the 2400 profit value. But, in OR-Library, the item which has 2400 profit value causes the resource violation. Whatever I do, the second chosen is being the item which has 2400 profit.
is there anything wrong my algorithm? I hope ppl who know somethings about ACO, should help me. Thanks in advance.
Input values:
6 10 3800//no of items (n) / no of resources (m) // the optimal value
100 600 1200 2400 500 2000//profits of items (6)
8 12 13 64 22 41//resource constraints matrix (m*n)
8 12 13 75 22 41
3 6 4 18 6 4
5 10 8 32 6 12
5 13 8 42 6 20
5 13 8 48 6 20
0 0 0 0 8 0
3 0 4 0 8 0
3 2 4 0 8 4
3 2 4 8 8 4
80 96 20 36 44 48 10 18 22 24//resource capacities.
My algorithm:
for i=0 to max_ant
for j=0; to item_number
if j==0
{
item=rand()%n
ant[i].value+=profit[item]
ant[i].visited[j]=item
}
else
{
calculate probabilities for all the other items in P[0..n]
find the biggest P value.
item=biggest P's item.
check if it is in visited list
check if it causes resource constraint.
if everthing is ok:
ant[i].value+=profit[item]
ant[i].visited[j]=item
}//end of else
}//next j
update pheremon matrix => tau[a][b]=rou*tau[a][b]+deltaTou
}//next i
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(?:$|0$))