UVA 1149

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Description

A set of n 1-dimensional items have to be packed in identical bins. All bins have exactly the same lengthl and each item i has lengthl_i $\le$ l . We look for a minimal number of binsq 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 , ...,l_n , to compute the optimal number of binsq .

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 line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The first line of the input file contains the number of items n(1 $\le$ n $\le$ 10₅) . The second line contains one integer that corresponds to the bin length l $\le$ 10000 . We then haven 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 two consecutive cases will be separated by a blank line.

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

Sample Input

1108070153035108020351030

Sample Output

Note: 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.

$\epsfbox{p3503.eps}$

题目比较好看懂

每个箱子最多能装两个

最少要几个箱子

排序一下

最左最右相加

如果大于length

最后单独一个箱子

right--

如果不大于

两个一个箱子

right-- left++

#include <stdio.h>#include <stdlib.h>#define  N 100010int item[N], length, n;int cmp(const void *a, const void *b) {return (*(int *)a - *(int *)b);}int main(){int c;scanf("%d", &c);getchar();while (c--) {scanf("%d", &n);scanf("%d", &length);int count = 0;for (int i = 0; i < n; i++)scanf("%d", &item[i]);qsort(item, n, sizeof(item[0]), cmp);int left = 0, right = n - 1;while (left <= right) {if (left == right) {count++;break;}if ((item[left] + item[right]) > length) {right--;}else {left++, right--;}count++;}printf("%d\n", count);if (c)printf("\n");}return 0;}

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