POJ2782（贪心）

Description
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 .
Input
The first line of the input contains the number of items n (1<=n<=10 5) . 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.
Output
Your program has to write the minimal number of bins required to pack all items.
Sample Input
10
80
70
15
30
35
10
80
20
35
10
30
Sample Output
6

贪心题，把长度依次从大到小排列，先把最大最小装一盒试试，如果可以，就把n-1，相当于把小盒和大盒装到了一起，否则大盒就只能单独装，依次装，直到装完。

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int maxn = 100000 + 5;int cmp(int a, int b){    return a > b;}int main(){    int n, l, li[maxn] = {0}, nu = 0;    cin >> n >> l;    for (int i = 0; i < n; i++) {        cin >> li[i];    }    sort (li, li + n, cmp);    for (int i = 0; i < n; i++) {        nu++;        if (li[i] + li[n - 1] <= l) n--;    }    cout << nu << endl;    return 0;}