M - Sum

最新推荐文章于 2019-10-02 17:32:18 发布

转载最新推荐文章于 2019-10-02 17:32:18 发布 · 92 阅读

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原文链接：http://www.cnblogs.com/wzl19981116/p/9347085.html

本文介绍了一种利用容斥原理解决特定数论问题的方法。该问题涉及到一系列操作，包括求解一定范围内与某个数互质的所有数之和，以及替换序列中的数值。文章通过实例演示了如何高效地处理这些操作。

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XXX is puzzled with the question below:

1, 2, 3, ..., n (1<=n<=400000) are placed in a line. There are m (1<=m<=1000) operations of two kinds.

Operation 1: among the x-th number to the y-th number (inclusive), get the sum of the numbers which are co-prime with p( 1 <=p <= 400000).
Operation 2: change the x-th number to c( 1 <=c <= 400000).

For each operation, XXX will spend a lot of time to treat it. So he wants to ask you to help him.

Input

There are several test cases.
The first line in the input is an integer indicating the number of test cases.
For each case, the first line begins with two integers --- the above mentioned n and m.
Each the following m lines contains an operation.
Operation 1 is in this format: "1 x y p".
Operation 2 is in this format: "2 x c".

Output

For each operation 1, output a single integer in one line representing the result.

Sample Input

1
3 3
2 2 3
1 1 3 4
1 2 3 6

Sample Output

7
0

用容斥原理求出不是互质的的数的和，在用等差求和减去不是互质的和，在对改动的数字进行特判

转载于:https://www.cnblogs.com/wzl19981116/p/9347085.html