题目：2234. 花园的最大总美丽值 - 力扣（LeetCode）

1. 先对flowers进行升序排序，计算现有“完善花园”的数量minFull；如果没有，minFull为n。

2. 计算前缀和f[]。

3. 从i=(minFull-1 to 0)枚举从第i个开始的花园都变成“完善花园”后剩余的花的数量，计算剩余的花可以使最小值最大的值

4. 因为我们是倒序枚举的，设“非完善花园”的最小数量是flowers[j]，则j在美剧过程中一定不会增长，可以通过f[]和剩余的花的数量快速计算出是否可以满足“非完善花园”的最小数量是flowers[j]这一条件；如果满足该条件后花朵数量还有剩余，可以计算可以给所有“非完善花园”各增加多少朵花

O(nlog(n))的排序+O(n)的遍历就可以得到答案了：

class Solution {
public:
    long long maximumBeauty(vector<int>& flowers, long long newFlowers, int target1, int full1, int partial1) {
        long long target = target1;
        long long full = full1;
        long long partial = partial1;
        
        sort(flowers.begin(), flowers.end());
        long long n = flowers.size();
        long long* f = new long long[n];
        if (flowers[0] >= target) {
            return full * n;
        }
        
        f[0] = flowers[0];
        long long minFull = n;
        for (int i = 1; i < n; i++) {
            f[i] = flowers[i];
            f[i] += f[i - 1];
            if (flowers[i] >= target) {
                minFull = i;
                break;
            }
        }
        long long ret = 0;
        long long j = minFull - 1;
        long long flowerJ = flowers[j];
        long long score, remain, temp;
        for (long long i = minFull - 1; i >= 0; i--) {
            if (j > i) {
                j = i;
                flowerJ = flowers[j];
            }
            while (j >= 0 && f[j] + newFlowers < flowerJ * (j + 1)) {
                j--;
                flowerJ = flowers[j];
            }
            remain = newFlowers - (flowerJ * (j + 1) - f[j]);
            temp = flowerJ + remain / (j + 1);
            if (temp >= target) {
                temp = target - 1;
            }
            score = (n - 1 - i) * full + temp * partial;
            if (score > ret) {
                ret = score;
            }
            
            newFlowers -= (target - flowers[i]);
            if (newFlowers < 0) {
                break;
            }
        }
        if (newFlowers >= 0) {
            score = n * full;
            if (score > ret) {
                ret = score;
            }
        }
        return ret;
    }
};