Nyoj-71-canoe trip

Trip on a canoe Time limit:3000MS | memory limit:65535KB Difficulty: 2

Description

During a canoe trip, the canoe can be rented at the port and there is no difference between them. A canoe can only take two people at most, and the total weight of a passenger cannot exceed the maximum carrying capacity of the canoe. We need to minimize the cost of this event, so we need to find the minimum number of canoes that can accommodate all passengers. WriteProgramThe maximum carrying capacity, number of passengers, and weight of each passenger. Based on the given rules, calculate the minimum number of canoes required for the placement of all passengers and output the results.

Input

Input s in the first line, indicating the number of test data groups;
The first row of each group of data contains two integers: W, N, 80 <= W <= 300, 1 <= n <=, W indicates the maximum carrying capacity of a canoe, and N indicates the number of people;
The next set of data is the weight of each person (not greater than the carrying capacity of the ship );

Output

The minimum number of canoes required for each group.

Sample Input

385 65 84 85 80 84 8390 390 45 60100 550 50 90 40 60

Sample output

 
533
 

Analysis: first, we should sort by size from small to large. After sorting, we thought that if the smallest weight cannot be the same as the largest weight, then the heavyweight will have to take a boat by themselves. This question can simulate this idea and constantly screen outSit-togetherOnly one person can sit in a few pairs and remove them. That is, after sorting, let the smallest and largest addition compare with the ship weight limit. If possible, both sides will be reduced by 1; otherwise, right minus 1, and sum ++; then, and then, the answer will come out.

DetailsCodeAs follows:

 
# Include <stdio. h ># include <algorithm> using namespace STD; int main () {int t, n, m, I, sum, J, s [301]; scanf ("% d", & T); While (t --) {scanf ("% d", & M, & N); for (I = 0; I <n; I ++) scanf ("% d", & S [I]); sort (S, S + n); for (I = 0, j = n-1, sum = 0;) {If (j <I) break; If (j = I) {sum ++; break ;} if (s [I] + s [J]> m) {sum ++; j --; continue;} If (s [I] + s [J] <= m) {sum ++; I ++; j --; Continue ;}} printf ("% d \ n", sum );}}