private readonly FooType _foo = new FooType();
/// <summary>
/// Gets the controller of this component
/// </summary>
private FooType Foo
{
get
{
return _foo;
}
}
@Override
public String toString() {
List<Integer> accountLevels = new ArrayList<>();
List<String> accountIds = new ArrayList<>();
for (SelectedHierarchyLevel selectedHierarchyLevel : selectedHierarchyLevels) {
accountLevels.add(selectedHierarchyLevel.getLevel());
accountIds.add(selectedHierarchyLevel.getAccountRowId());
}
vetoAreValidIds(accountIds);
StringBuilder converterString = new StringBuilder();
convert(accountLevels, converterString);
converterString.append("$");
convert(accountIds, converterString);
return converterString.toString();
}
Just because you can, doesn’t mean you should.
var list = new List<string>() { // Assume some items in the list };
for (var i = 0; i < list.Count; i++)
{
var item = list[i];
list.Remove(item);
i--;
}
Simple alternative to List.Clear()
roots.push(resolver(Config.root, `${pageDir}${path.sep}${page}${path.sep}${page}.html`));
scrollToEl($el.parent().parent().parent());
Someones idea of how to select an element to scroll to.... Please no.
public class Main {
public static void main(String[] args) {
try {
//code goes here
} catch (Exception e) {
System.exit(0);
}
}
}
public static Integer find(List<String> list, String name, int i) {
if(list.get(i).equals(name)) {
return i;
}
else return find(list, name, i+1)
}
if ((&inactiveSlot)->GetFirstSlot() == RuntimeLib::INVALID_SLOT) {
Dot dude...
if (CATEGORY_NORMAL.equalsIgnoreCase(categorie)) {
return assignGroupStartWithPrefix(assignmentGroups);
} else if (CATEGORY_EXTERNAL.equalsIgnoreCase(categorie)) {
return assignGroupStartWithPrefix(assignmentGroups);
}
#include <stdio.h>
#include<stdlib.h>
#include<time.h>
void randy(int array[] ){
int i = 0 , d = 0;
for(i = 1 ; i <= 23 ; i++){
array[i] = 1 + (rand() % 365 );
}
}
int compare (const void * a, const void * b)
{
return ( *(int*)a - *(int*)b );
}
int find (int arr[]){
int d = 0;
for(d = 1 ;d <=23 ; d++ ){
if(arr[d] == arr[d + 1]){
return 1;
}
}
return 0;
}
int main(void){
double count_birthday = 0;
int people[24];
srand(time(NULL));
int c = 0 ;
long double ip = 0;
while(ip++ <= 1000000 ){
randy(people);
qsort (people, 25, sizeof(int), compare);
if(find (people) == 1 ){
count_birthday++;
}
}
printf("%.2lf", count_birthday / 10000 );
return 0;
}
namespace network
{
class ip
{
uint _IP;
public ushort this[int i]
{
get
{
switch(i)
{
case 0:
case 1:
case 2:
case 3:
return (ushort)(_IP>>(i*8));
default:
throw new IndexOutOfRangeException();
}
}
set
{
switch(i)
{
case 0:
case 1:
case 2:
case 3:
_IP=(((uint)value)<<i*8);
break;
default:
throw new IndexOutOfRangeException();
}
}
}
}
}
I have no idea what this does
# 5-level loop, forgive me...
for xi, xs in enumerate(X):
for yi, ys in enumerate(Y):
for zi, zs in enumerate(Z):
lx, ly, lz = len(xs), len(ys), len(zs)
# construct points
xx, yy, zz = custom_meshgrid(xs, ys, zs)
world_xyzs = (
torch.cat(
[xx.reshape(-1, 1), yy.reshape(-1, 1), zz.reshape(-1, 1)],
dim=-1,
)
.unsqueeze(0)
.to(count.device)
) # [1, N, 3]
# cascading
for cas in range(self.cascade):
bound = min(2**cas, self.bound)
half_grid_size = bound / resolution
# scale to current cascade's resolution
cas_world_xyzs = world_xyzs * (bound - half_grid_size)
# split batch to avoid OOM
head = 0
while head < B:
tail = min(head + S, B)
# world2cam transform (poses is c2w, so we need to transpose it. Another transpose is needed for batched matmul, so the final form is without transpose.)
cam_xyzs = cas_world_xyzs - poses[
head:tail, :3, 3
].unsqueeze(1)
cam_xyzs = cam_xyzs @ poses[head:tail, :3, :3] # [S, N, 3]
# query if point is covered by any camera
mask_z = cam_xyzs[:, :, 2] > 0 # [S, N]
mask_x = (
torch.abs(cam_xyzs[:, :, 0])
< cx / fx * cam_xyzs[:, :, 2] + half_grid_size * 2
)
mask_y = (
torch.abs(cam_xyzs[:, :, 1])
< cy / fy * cam_xyzs[:, :, 2] + half_grid_size * 2
)
mask = (
(mask_z & mask_x & mask_y).sum(0).reshape(lx, ly, lz)
) # [N] --> [lx, ly, lz]
# update count
count[
cas,
xi * S : xi * S + lx,
yi * S : yi * S + ly,
zi * S : zi * S + lz,
] += mask
head += S
// enum - full enumeration of knapsack solutions
// (C) Joshua Knowles
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
FILE *fp; // file pointer for reading the input files
int Capacity; // capacity of the knapsack (total weight that can be stored)
int Nitems; // number of items available
int *item_weights; // vector of item weights
int *item_values; // vector of item profits or values
int *temp_indexes; // list of temporary item indexes for sorting items by value/weight
int QUIET=0; // this can be set to 1 to suppress output
extern void read_knapsack_instance(char *filename);
extern void print_instance();
extern void sort_by_ratio();
extern int check_evaluate_and_print_sol(int *sol, int *total_value, int *total_weight);
void enumerate();
int next_binary(int *str, int Nitems);
int main(int argc, char *argv[])
{
read_knapsack_instance(argv[1]);
print_instance();
enumerate();
return(0);
}
void enumerate()
{
// Do an exhaustive search (aka enumeration) of all possible ways to pack
// the knapsack.
// This is achieved by creating every binary solution vector of length Nitems.
// For each solution vector, its value and weight is calculated.
int i; // item index
int solution[Nitems+1]; // (binary) solution vector representing items packed
int best_solution[Nitems+1]; // (binary) solution vector for best solution found
int best_value; // total value packed in the best solution
double j=0;
int total_value, total_weight; // total value and total weight of current knapsack solution
int infeasible; // 0 means feasible; -1 means infeasible (violates the capacity constraint)
// set the knapsack initially empty
for(i=1;i<=Nitems;i++)
{
solution[i]=0;
}
QUIET=1;
best_value=0;
while(!(next_binary(&solution[1], Nitems)))
{
/* ADD CODE IN HERE TO KEEP TRACK OF FRACTION OF ENUMERATION DONE */
// calculates the value and weight and feasibility:
infeasible=check_evaluate_and_print_sol(solution, &total_value, &total_weight);
/* ADD CODE IN HERE TO KEEP TRACK OF BEST SOLUTION FOUND*/
}
/* ADD CODE TO PRINT OUT BEST SOLUTION */
}
int next_binary(int *str, int Nitems)
{
// Called with a binary string of length Nitems, this
// function adds "1" to the string, e.g. 0001 would turn to 0010.
// If the string overflows, then the function returns 1, else it returns 0.
int i=Nitems-1;
while(i>=0)
{
if(str[i]==1)
{
str[i]=0;
i--;
}
else
{
str[i]=1;
break;
}
}
if(i==-1)
{
return(1);
}
else
{
return(0);
}
}
A genuine UoM lab.
public static class time
{
public const int minutes_per_hour = 60;
public const int seconds_per_minute = 60;
public const int hours_per_day = 24;
public const int minutes_per_day = minutes_per_hour * hours_per_day;
public const int seconds_per_day = seconds_per_minute * minutes_per_day;
public const int days_per_year = 365;
}
nohup find dump_events/* -type f -name "*.jpg" -exec bash -c "cp -v {} \"backup_events/\`ls {} --full-time --time-style="+%Y-%m-%d_%H-%M-%S" | cut -d ' ' -f 6 \`_\`md5sum {} | cut -d ' ' -f 1\`.jpg\";" \; &
Let's do this... I may also add this... And this... And that... And the other thing there... Then calculate how flat the earth is... But in one line!