題目鏈接：

題目：

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.

思路：

啊 沒想到這次contest中 我遇到最難的題是這道。。。用F(k)=F(k-1)-(n-1)*end+(sum-end)  + 0*end = F(k-1)+sum-n*end

畫了個圖：

算法：

[java] view plain copy

 

1. public int maxRotateFunction(int[] A) {  
2.     int max=Integer.MIN_VALUE;  
3.     int sum = 0;  
4.     int pre = 0;  
5.       
6.     for(int i=0;i<A.length;i++){  
7.         pre +=A[i]*i;  
8.         sum+=A[i];  
9.     }  
10.     max = Math.max(pre, max);  
11.   
12.     int k=1;  
13.     while(k<A.length){  
14.         int res = pre+sum-A.length*A[A.length-k];  
15.         max = Math.max(max, res);  
16.         pre = res;  
17.         k++;  
18.     }  
19.       
20.     return max;  
21. }