【hihocoder1051】补提交卡——贪心

题目：点击进入

描述：

一个1到100的数列有n个地方是断开的，现给出这n个位置以及一个数m，表示可以使m个位置连续，求可以得到的最长的连续区间的长度，n个位置按照升序给出。

题解：

将起点和终点补充进去之后，必须使这m个位置连续地消除才能使最终长度最长，所以直接枚举这m个位置的左端点即可。

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <climits>
#include <cassert>
#include <cctype>
#include <complex>
#include <algorithm>
#include <string>
#include <iostream>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <set>
using namespace std;
#define fi first
#define se second
#define MP(A, B) make_pair(A, B)
#define pb push_back
#define gcd __gcd
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
typedef long long ll;
typedef unsigned long long ulls;
typedef unsigned int uint;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vii;
typedef map<int, int> mii;
typedef map<string, int> msi;
typedef map<pii, int> mpi;
#if ( ( _WIN32 || __WIN32__ ) && __cplusplus < 201103L)
    #define lld %I64d
#else
    #define lld %lld
#endif
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;
const int maxn = 1e2 + 5;
const int maxm = 1e6 + 5;
const int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
inline int scan(int &a) { return scanf("%d", &a); }
inline int scan(int &a, int &b) { return scanf("%d%d", &a, &b); }
inline int scan(int &a, int &b, int &c) { return scanf("%d%d%d", &a, &b, &c); }
inline int scan(ll &a) { return scanf("lld", &a); }
inline int scan(ll &a, ll &b) { return scanf("lldlld", &a, &b); }
inline int scan(ll &a, ll &b, ll &c) { return scanf("lldlldlld", &a, &b, &c); }
inline int scan(double &a) { return scanf("%lf", &a); }
inline int scan(double &a, double &b) { return scanf("%lf%lf", &a, &b); }
inline int scan(double &a, double &b, double &c) { return scanf("%lf%lf%lf", &a, &b, &c); }
inline int scan(char &a) { return scanf("%c", &a); }
inline int scan(char *a) { return scanf("%s", a); }
template<class T> inline void mem(T &A, int x) { memset(A, x, sizeof(A)); }
template<class T0, class T1> inline void mem(T0 &A0, T1 &A1, int x) { mem(A0, x), mem(A1, x); }
template<class T0, class T1, class T2> inline void mem(T0 &A0, T1 &A1, T2 &A2, int x) { mem(A0, x), mem(A1, x), mem(A2, x); }
template<class T0, class T1, class T2, class T3> inline void mem(T0 &A0, T1 &A1, T2 &A2, T3 &A3, int x) { mem(A0, x), mem(A1, x), mem(A2, x), mem(A3, x); }
template<class T0, class T1, class T2, class T3, class T4> inline void mem(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, int x) { mem(A0, x), mem(A1, x), mem(A2, x), mem(A3, x), mem(A4, x); }
template<class T0, class T1, class T2, class T3, class T4, class T5> inline void mem(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, int x) { mem(A0, x), mem(A1, x), mem(A2, x), mem(A3, x), mem(A4, x), mem(A5, x); }
template<class T0, class T1, class T2, class T3, class T4, class T5, class T6> inline void mem(T0 &A0, T1 &A1, T2 &A2, T3 &A3, T4 &A4, T5 &A5, T6 &A6, int x) { mem(A0, x), mem(A1, x), mem(A2, x), mem(A3, x), mem(A4, x), mem(A5, x), mem(A6, x); }
template<class T> inline T min(T a, T b, T c) { return min(min(a, b), c); }
template<class T> inline T max(T a, T b, T c) { return max(max(a, b), c); }
template<class T> inline T min(T a, T b, T c, T d) { return min(min(a, b), min(c, d)); }
template<class T> inline T max(T a, T b, T c, T d) { return max(max(a, b), max(c, d)); }
template<class T> inline T min(T a, T b, T c, T d, T e) { return min(min(min(a,b),min(c,d)),e); }
template<class T> inline T max(T a, T b, T c, T d, T e) { return max(max(max(a,b),max(c,d)),e); }
int a[maxn];
int main()
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
        long _begin_time = clock();
    #endif

    int T; scan(T);
    while(T--)
    {
        int n, m; scan(n, m);
        for(int i = 1; i <= n; i++) scan(a[i]);
        if(m >= n)
        {
            printf("100\n");
            continue;
        }
        a[0] = 0, a[(++n)++] = 101;
        int ans = 0;
        for(int i = 0; i < n; i++)
        {
            if(i + m + 1 > n) break;
            ans = max(ans, a[i + m + 1] - a[i] - 1);
        }
        printf("%d\n", ans);
    }

    #ifndef ONLINE_JUDGE
        long _end_time = clock();
        printf("time = %ld ms\n", _end_time - _begin_time);
    #endif

    return 0;
}