Data structure - Fenwick tree - Modify a single point interval evaluation interval plus -1750 and evaluation

2020-04-03 12:35:23

Problem Description:

A known number of columns, you need to perform the following two operations:
1. a number of sections each plus X
2. obtaining a number of value
inputs:
the original array is A. For convenience, A [0] is 0. The actual number of columns from the A [1] starts.
Operation given by the 4-tuple.
For each 4-tuple (a, b, c, d ):
If a = 0 requires A [b] -A [c] values have increased interval D (modified).
If required to give a value a = 1 A [b] a. Wherein c, d inoperative (query).

Output:
To reduce the amount of output data. 2 operation (query) Returns the value of all the exclusive OR (^).

Sample

Example 1:

输入：[0,1,2,3,4],[[1,1,0,0],[0,1,2,1],[1,2,0,0]]
输出：2
解释：
第一个操作返回A [1] = 1
第二个操作改变A为 [0,2,3,3,4]
第三个操作返回A [1] = 3
所以 1 ^ 3 = 2

Sample 2:

输入：[0,1],[[1,1,0,0]]
输出：1
解释：第一个操作返回A [1] = 1,答案为 1。

Precautions

• Sequence length <= 10000
• Operand <= 50000

Problem Solving:

    int[] bit;
    public long intervalsAddAndGetValue(int[] A, int[][] operations) {
        int n = A.length;
        bit = new int[n];
        for (int i = 1; i < n; i++) {
            update(i, A[i]);
            update(i + 1, -A[i]);
        }
        long res = 0;
        for (int[] op : operations) {
            if (op[0] == 0) {
                update(op[1], op[3]);
                update(op[2] + 1, -op[3]);
            }
            else {
                res ^= query(op[1]);
            }
        }
        return res;
    }
    
    private void update(int idx, int delta) {
        for (int i = idx; i < bit.length; i += (i & -i)) {
            bit[i] += delta;
        }
    }
    
    private int query(int idx) {
        int res = 0;
        for (int i = idx; i > 0; i -= (i & -i)) {
            res += bit[i];
        }
        return res;
    }