[HIHO1051]补提交卡（枚举，贪心）

题目链接：http://hihocoder.com/problemset/problem/1051

思路：先排序，然后枚举连续的长度为m的子段，用这个段之后的第一个天数减去这个段之前的第一个天数再-1就是把这个区间全填上后的连续提交天数。

 /*
 ━━━━━┒ギリギリ♂ eye！
 ┓┏┓┏┓┃キリキリ♂ mind！
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┃┃┃┃┃┃
 ┻┻┻┻┻┻
 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%I64d", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(LL i = 0; i < (len); i++)
 #define For(i, a, len) for(LL i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Fuint(a) memset((a), 0x7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long Uint;
 typedef pair<LL, LL> pii;
 typedef pair<string, LL> psi;
 typedef map<string, LL> msi;
 typedef vector<LL> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 const int maxn = ;
 int n, m;
 int a[maxn];
 int G[maxn][maxn];

 int main() {
     // FRead();
     int T;
     Rint(T);
     W(T) {
         Cls(G); Cls(a);
         Rint(n); Rint(m);
         For(i, , n+) Rint(a[i]);
         if(m >= n) {
             printf("100\n");
             continue;
         }
         sort(a+, a+n+);
         int ret = ;
         for(int i = ; i <= n - m + ; i++) {
             int j = i + m - ;
             int tmp = a[j+] - a[i-] - ;
             ret = max(ret, tmp);
         }
         printf("%d\n", ret);
     }
     RT ;
 }