You already mentioned the solution in your question. You can use: range(0, 10) To repeat the sequence 3 times you could do: str_repeat(implode(‘ ‘,range(0, 10)).’ ‘, 3); 1 solved How to range and sequence in PHP
This works for me. public static void main(final String[] args) { final String input = “aaaab aOa baaab c”; final String[] sections = input.split(” “); final int length = 3; final List<String> list = new ArrayList<>(); for (final String section : sections) { for (int i = 0; i < section.length(); i++) { if (section.length() … Read more
For Sql Sever Select Row_Number() Over(Order By [EMP_NO] Desc) as Emp_No, EMP_NAME from TableName Order By [Emp_No] Desc Sql Fiddle Demo For Oracle Sql Developer Select Row_Number() Over(Order By “EMP_NO” Desc) as “Emp_No”, “EMP_NAME” from Table1 Order By “Emp_No” Desc Sql Fiddle Demo 18 solved how to reverse a sequence [closed]
There is a Sequence associated with the table. When i reset this sequence and inserting a data into a table data is not coming in ordered fashion solved There is a Sequence associated with the table. When i reset this sequence and inserting a data into a table data is not coming in ordered fashion
Yes, it is possible to find modulo (p) and sequence value (y) for given number x, but it requires using floating-arithmetics logarithm. pow = Floor(Log2(x+1)) p = 1 << pow y = 1 + (x + 1) % p For initial period s (3 in your second example) pow = Floor(Log2((x + s – 1) … Read more
Many options here. Unfortunately you don’t show any effort to resolve the problem or even to make it clear and reproducible. For example, using ifelse : dat <- read.table(text=””aaa” 28 “aac” 8 “aag” 20″) dat$V1 <- with(dat,ifelse(V1 %in% c(‘aaa’,’aag’),’K’,’N’)) dat V1 V2 1 K 28 2 N 8 3 K 20 EDIT If I want … Read more
try this one: import java.io.BufferedReader; import java.io.InputStreamReader; public class Rozwiazanie { public static void main(String[] args) { try { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); String[] splittedLinia = br.readLine().split((char) 32 + “”);//moglaby byc ” ” ale tak na wszelki wypadek nie ma chuja zeby sie popierdolilo teraz nawet na linuxie int aktualnyWyraz = Integer.parseInt(splittedLinia[0]);//uwaga jakby … Read more
When using combinations() you are getting all pairs, even the non-adjacent ones. You can create a function that will return the adjacent pairs. I’ve tried the following code and it worked, maybe it can give you some insight: import os import re from collections import Counter def pairs(text): ans = re.findall(r'[A-Za-z]+’, text) return (tuple(ans[i:i+2]) for … Read more
You can use a sequence comprehension: let values = seq { for i in 1110 .. 10 .. 6660 do for j in 1 .. 6 do yield i + j } and create a set using Set.ofSeq e.g. let s = Set.ofSeq values 1 solved Creating Set in F# with elements from 1111 to … Read more
#include <stdio.h> int seq(void){ static int v = 5, d = 1; int ret = v; v += d; if(v == 5 || v == 7) d = -d; return ret; } int main(void) { for(int i = 0; i < 20; ++i) printf(“%d “, seq()); return 0; } 1 solved How to iterate over … Read more
There are many possible solutions to this problem. Here’s a reusable solution that simply decrements from 4 to 1 and adds the expected number of 1’s and 0’s. Loops used : 1 def sequence(n): string = “” for i in range(n): string+=’1’*(n-i) string+=’0’*(n-i) return string print sequence(4) There’s another single-line elegant and more pythonic way … Read more