396. Rotate Function ------------M

 Given an array of integers A and let n to be its length.

Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow:

F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1].

Calculate the maximum value of F(0), F(1), ..., F(n-1).

Note:
n is guaranteed to be less than 105.

Example:

A = [4, 3, 2, 6]

F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25
F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16
F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23
F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26

So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.

A:

找到通项公式。

并计算即可

class Solution {
public:
    int maxRotateFunction(vector<int>& A) {
        long total = accumulate(A.begin(), A.end(), 0L);
        
        int n = A.size();
        long F0 = 0;
        for(long i =0;i<n;i++)
            F0 += i * A[i];
        // F(k) = F0 + total * k - n * ( A[n-1] + A[n-2] + ....A[n-k] )
        long preAkSum = 0; // this stands for ( A[n-1] + A[n-2] + ....A[n-k] )
        long resMax = F0;
        for(int k=1;k<=n-1;k++){
            preAkSum += A[n-k];
            long Fk = F0 + total * k - n * preAkSum;
            resMax = max(resMax, Fk);
        }
        return (int)resMax;
    }
};