Today, the bookstore owner has a store open for customers.length minutes. Every minute, some number of customers (customers[i]) enter the store, and all those customers leave after the end of that minute.
On some minutes, the bookstore owner is grumpy. If the bookstore owner is grumpy on the i-th minute, grumpy[i] = 1, otherwise grumpy[i] = 0. When the bookstore owner is grumpy, the customers of that minute are not satisfied, otherwise they are satisfied.
The bookstore owner knows a secret technique to keep themselves not grumpy for X minutes straight, but can only use it once.
Return the maximum number of customers that can be satisfied throughout the day.
Example 1:
Input: customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], X = 3 Output: 16 Explanation: The bookstore owner keeps themselves not grumpy for the last 3 minutes. The maximum number of customers that can be satisfied = 1 + 1 + 1 + 1 + 7 + 5 = 16.
Note:
1 <= X <= customers.length == grumpy.length <= 200000 <= customers[i] <= 10000 <= grumpy[i] <= 1题目大意：
customers表示这个时间段内会有多少个客户在商店内，grumpy表示老板是否生气如果生气则客户的就会不满意，但是给出一个X表示可以在这个时间段内变得不生气，求解最多会有多少顾客满意。
解题思路：
暴力即可。
若X大于整个的时间段，则我们将输出全部客户的满意度。
否组，先对原有的customers数组进行处理，将boss生气的部分置为负数，统计此时最大的满意度。之后遍历数组，计算X区间内最大满意度加上原有不满意的部分，最终求解最大值。
class Solution { public: int maxSatisfied(vector<int>& c, vector<int>& g, int X) { int n = c.size(); int sum_c_good = 0; int sum_c_bad = 0; for(int i = 0;i<n;i++){ sum_c_good += c[i]; if(g[i]==1){ c[i] *= -1; }else{ sum_c_bad += c[i]; } } int ans = sum_c_bad; if(X>=n){ ans = sum_c_good; }else{ for(int i=0;i<=n-X;i++){ int tmp_idx = i; int cor_sum = 0; int tmp_sum_c_bad = sum_c_bad; for(int j=0;j<X;j++){ if(c[tmp_idx+j]<0){ tmp_sum_c_bad += abs(c[tmp_idx+j]); } } ans = max(tmp_sum_c_bad, ans); } } return ans; } };
