其实这个题也可以二分，就是这个二段性全在数目上

C++ 代码

class Solution {
public:
    int maximumRobots(vector<int>& chargeTimes, vector<int>& runningCosts, long long budget) {
        int n = chargeTimes.size();
        int l = 0, r = 0;
        int ans = 0;
        long long sumT = 0;
        deque<int> q; 

        for (; r < n; r++) {
            while (!q.empty() && chargeTimes[q.back()] <= chargeTimes[r]) {
                q.pop_back();
            }
            q.push_back(r);
            sumT += runningCosts[r];

            while (l <= r && chargeTimes[q.front()] + (r - l + 1) * sumT > budget) {
                if (q.front() == l) {
                    q.pop_front();
                }
                sumT -= runningCosts[l++];
            }
            ans = max(ans, r - l + 1);
        }

        return ans;
    }
};