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; } };
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