牛客网暑期ACM多校训练营（第九场）H. Prefix Sum（CDQ分治）

题目描述

Niuniu has learned prefix sum and he found an interesting about prefix sum.

Let's consider (k+1) arrays a[i] (0 <= i <= k)

The index of a[i] starts from 1. 

a[i] is always the prefix sum of a[i-1]. 

"always" means a[i] will change when a[i-1] changes.

"prefix sum" means a[i][1] = a[i-1][1] and a[i][j] = a[i][j-1] + a[i-1][j] (j >= 2)

 

Initially, all elements in a[0] are 0.

There are two kinds of operations, which are modify and query.

For a modify operation, two integers x, y are given, and it means a[0][x] += y.

For a query operation, one integer x is given, and it means querying a[k][x].

As the result might be very large, you should output the result mod 1000000007.

输入描述:

The first line contains three integers, n, m, k.
n is the length of each array.
m is the number of operations.
k is the number of prefix sum.

In the following m lines, each line contains an operation.
If the first number is 0, then this is a change operation.
There will be two integers x, y after 0, which means a[0][x] += y;
If the first number is 1, then this is a query operation.
There will be one integer x after 1, which means querying a[k][x].

1 <= n <= 100000
1 <= m <= 100000
1 <= k <= 40
1 <= x <= n
0 <= y < 1000000007

输出描述:

For each query, you should output an integer, which is the result.

输入

4 11 3
0 1 1
0 3 1
1 1
1 2
1 3
1 4
0 3 1
1 1
1 2
1 3
1 4

输出

1
3
7
13
1
3
8
16

 

题意：给你一个(k+1)*n的矩阵，初始全为0，m个操作

操作①0 x y表示将a[0][x] += y，并且计算∀(i∈[1,k]，j∈[1,n])，a[i][j] = a[i-1][j]+a[i][j-1]

操作②1 x 表示询问a[k][x]的值

 

很容易算出：a[0][x]对a[k][y]的贡献为，因为k的范围很小所以可以O(1)预处理所有组合数

用上面的结论，直接暴力每个操作对后面所有询问的贡献，复杂度是O(n*m)的，肯定会超时

对于这种每次操作对后面询问产生特定贡献的题，考虑对询问时间进行CDQ分治

 

步骤如下：

1. 按照操作的时间进行排序（其实根本不用排，因为本来就有序，输入顺序就是了）
2. 对操作进行二分，假设当前二分区间为[L, R]，一个很显然的结论是对于[L, m]的所有修改操作，一定对(m, R]区间中的询问操作有效
3. 所以对于每个二分区间[L, R]，只需要计算[L, m]中的所有修改操作对(m, R]区间中所有询问操作的贡献即可
4. 如果区间长度len≤2000，直接暴力所有操作对所有询问的贡献，复杂度O(len²)
5. 如果区间长度len>2000，将所有操作全部加到矩阵中，然后暴力计算出矩阵所有位置的值，最后对于所有询问直接查矩阵对应位置（这里可以优化：不用暴力整个矩阵的值，只需要用组合数计算最后一排的值即可）复杂度O(nk)或O(n)
6. 分析下整体复杂度：len>2000的递归次数最坏情况下(n=m=100000)约为15次，所以这部分复杂度为O(15n)， len<2000的递归次数约为m次，每次复杂度均摊log²(m)次，所以这部分复杂度O(mlogm)，整体复杂度O(15n+mlog²m)
7. 搞定！

 

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