题目：
A set of n 1-dimensional items have to be packed in identical bins. All bins have exactly the same length l and each item i has length li<=l . We look for a minimal number of bins q such that
? each bin contains at most 2 items,
? each item is packed in one of the q bins,
? the sum of the lengths of the items packed in a bin does not exceed l .
You are requested, given the integer values n , l , l1 , …, ln , to compute the optimal number of bins q .
输入：
The first line of the input contains the number of items n (1<=n<=105). The second line contains one integer that corresponds to the bin length l<=10000. We then have n lines containing one integer value that represents the length of the items.
输出：
Your program has to write the minimal number of bins required to pack all items.
输入样例：
10
80
70
15
30
35
10
80
20
35
10
30
输出样例：
6
提示：
The sample instance and an optimal solution is shown in the figure below. Items are numbered from 1 to 10 according to the input order.
题意：
给你很多大小不一的物品，装到大小为L的盒子里，且每个盒子最多装两个物品，问最少需要多少个盒子
题解：
从小到大排序。
利用两个指针，小指针从小到大，大指针从大到小遍历，使用一个盒子，先取大那头物体装入，大指针左移，如果剩余空间足够再装个小的，则将小的装入，小指针右移，否则继续取盒子，重复到所有物体装完(指针相交错开)。
AC代码：
#include<iostream>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
int n,m,ans;
int a[100005];
int main(){
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
}
sort(a,a+n);
ans=0;
for(int i=0,j=n-1;i<=j;){
ans++;
if(a[i]+a[j]>m){
j--;
}
else{
i++;
j--;
}
}
printf("%d\n",ans);
return 0;
}