題目地址：

https://www.lintcode.com/problem/take-the-element-and-query-the-sum/description

給定一個長$n$的陣列$A$，求${\sum }_{0\le i<j<n}A[i]A[j]$。

思路是字首和陣列。可以這樣看：$$\sum _{0\le i<j<n}A[i]A[j]=\sum _{j=1}^{n-1}(A[j]\sum _{i=0}^{j-1}A[i])$$程式碼如下：

public class Solution {
    /**
     * @param arr: the arr
     * @return: the sum
     */
    public int takeTheElementAndQueryTheSum(int[] arr) {
        // Write your code here
        long[] preSum = new long[arr.length + 1];
        long res = 0, MOD = (long) (1E9 + 7);
        for (int i = 0; i < arr.length; i++) {
            preSum[i + 1] = preSum[i] + arr[i];
            preSum[i + 1] %= MOD;
        }
        
        for (int i = arr.length - 1; i >= 0; i--) {
            res += arr[i] * preSum[i] % MOD;
            res %= MOD;
        }
        
        return (int) res;
    }
}

時空複雜度$O(n)$。