ACM Boxing problem

Description

A set of n<tex2html_verbatim_mark> 1-dimensional items has A to is packed in identical bins. All bins has exactly the same length l<tex2html_verbatim_mark> and each item i<tex2ht Ml_verbatim_mark> has length lil<tex2html_verbatim_mark>. We look for a minimal number of bins q<tex2html_verbatim_mark> such that

• Each bin contains at the most 2 items,
• Each item was packed in one of the Q<tex2html_verbatim_mark> bins,
• The sum of the lengths of the items packed in a bin does not exceed l<tex2html_verbatim_mark>.

You are requested, given the integer values n<tex2html_verbatim_mark>, l<tex2html_ verbatim_mark>, l1<tex2html_verbatim_mark>, ..., ln<tex2html_verbatim_ Mark>, to compute the optimal number of bins q<tex2html_verbatim_mark>.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This was followed by a blank line, and there was also a blank line between the consecutive inputs.

The first line of the input file contains the number of items n<tex2html_verbatim_mark>(1n ) <tex2html_verbatim_mark>. The second line contains one integer so corresponds to the bin length l10000<tex2html_verbatim_mark> . We then had n<tex2html_verbatim_mark> lines containing one integer value that represents the length of The items.

Output

For each test case, the output must follow the description below. The outputs of the consecutive cases would be separated to a blank line.

For each input file, your program has a to write the minimal number of bins required to pack all items.

Sample Input

1108070153035108020351030

Sample Output

6

Problem Solving Ideas:
The main idea of this topic is that given the length of the block and the length of the box that can be used for the block, it is required that a box can fit up to two blocks, asking us to find out how many boxes we need to use. We first put the boxes in order from large to small, and then let a current largest box and a minimum box combination to see if it can be loaded into the box, if not, then we only loaded into the big box, otherwise the two boxes are loaded. (We can use X, y to separate control from left to right and right to left subscript) notice the blank line of the title, I just keep hanging on this empty line.
Program code:

#include <iostream>#include<algorithm>using namespacestd;inta[100005];intFunintAintBB) {    returnA>b;}intMain () {intC; CIN>>C; intg=1;  while(c--)    {        if(g++!=1) cout<<Endl; intn,l; CIN>>N; CIN>>l;  for(intI=0; i<n;i++) {cin>>A[i]; } sort (A,a+n,fun); intx=0; inty=n-1; ints=0;  while(x<y) {if(a[x]+a[y]<=l) {x++; Y--; S++; }            Else{x++; S++; }        }         if(x==y) s++; cout<<s<<Endl; }    return 0;}

ACM Boxing problem