CF#196DIV2:A-Puzzles

http://codeforces.com/contest/337/problem/A

The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for hern students and give each of them a jigsaw puzzle
(which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).

The shop assistant told the teacher that there are m puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists off_{1
pieces, the second one consists off2 pieces and so on.}

Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. LetA be the number of pieces in the largest puzzle that the
teacher buys andB be the number of pieces in the smallest such puzzle. She wants to choose suchn puzzles that
A - B is minimum possible. Help the teacher and find the least possible value ofA - B.

Input

The first line contains space-separated integers n andm (2 ≤ n ≤ m ≤ 50). The second line containsm
space-separated integers f_{1, f2, ..., fm (4 ≤ fi ≤ 1000)
— the quantities of pieces in the puzzles sold in the shop.}

Output

Print a single integer — the least possible difference the teacher can obtain.

Sample test(s)Input

4 610 12 10 7 5 22

Output

5

Note

Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5.
It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.

 

題意：給出n,m,n是學生數，m是禮物數，在m哥禮物中 找出n個禮物，是的最大與最小值得差最小

思路：由於資料小，直接排序暴力

 

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int main(){    int n,m,i,a[55],Min,s;    while(~scanf("%d%d",&m,&n))    {        for(i = 0; i<n; i++)            scanf("%d",&a[i]);        sort(a,a+n);        Min = 10000;        m = m-1;        for(i = m; i<n; i++)        {            s = a[i]-a[i-m];            if(s<Min)                Min = s;        }        printf("%d\n",Min);    }    return 0;}