hihocoder1051(枚举贪心)

题目链接：点击打开链接

解题思路：

明显要消除连续的m才能使收益最大，我们直接暴力的枚举好了，每个区间的a[i + m ] - a[ i - 1] - 1的最大值即所求。这里在左右边界分别添上0和100

完整代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>
using namespace std;
int n , m;
const int maxn = 100001;
const int INF = 1000000000;
int a[maxn];
void solve()
{
    int maxx = -INF;
    for(int i = 1 ; i <= n ; i ++)
    {
        if(i + m > n)   break;
        int sum = a[i + m] - a[i - 1] - 1;
        if(sum > maxx)
            maxx = sum;
    }
    cout << maxx << endl;
}

int main()
{
    #ifdef DoubleQ
    freopen("in.txt" , "r" , stdin);
    #endif // DoubleQ
    int T;
    cin >> T;
    while(T--)
    {
        cin >> n >> m;
        for(int i = 1 ; i <= n ; i ++)
            cin >> a[i];
        if(m >= n)
        {
            cout << "100" << endl;
            continue;
        }
        a[0] = 0;
        a[n+1] = 100;
        n ++;
        solve();
    }
}