A15334. Madoka and The Best University
编程题
普及/提高-
知识点
题目描述
Madoka wants to enter to "Novosibirsk State University", but in the entrance exam she came across a very difficult task:
Given an integer $n$ , it is required to calculate $\sum{\operatorname{lcm}(c, \gcd(a, b))}$ , for all triples of positive integers $(a, b, c)$ , where $a + b + c = n$ .
In this problem $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $x$ and $y$ , and $\operatorname{lcm}(x, y)$ denotes the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of $x$ and $y$ .
Solve this problem for Madoka and help her to enter to the best university!
Given an integer $n$ , it is required to calculate $\sum{\operatorname{lcm}(c, \gcd(a, b))}$ , for all triples of positive integers $(a, b, c)$ , where $a + b + c = n$ .
In this problem $\gcd(x, y)$ denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor) of $x$ and $y$ , and $\operatorname{lcm}(x, y)$ denotes the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of $x$ and $y$ .
Solve this problem for Madoka and help her to enter to the best university!
输入格式
The first and the only line contains a single integer $n$ ( $3 \le n \le 10^5$ ).
输出格式
Print exactly one interger — $\sum{\operatorname{lcm}(c, \gcd(a, b))}$ . Since the answer can be very large, then output it modulo $10^9 + 7$ .
输入输出样例
输入 #1
3
输出 #1
1
输入 #2
5
输出 #2
11
输入 #3
69228
输出 #3
778304278
说明/提示
In the first example, there is only one suitable triple $(1, 1, 1)$ . So the answer is $\operatorname{lcm}(1, \gcd(1, 1)) = \operatorname{lcm}(1, 1) = 1$ .
In the second example, $\operatorname{lcm}(1, \gcd(3, 1)) + \operatorname{lcm}(1, \gcd(2, 2)) + \operatorname{lcm}(1, \gcd(1, 3)) + \operatorname{lcm}(2, \gcd(2, 1)) + \operatorname{lcm}(2, \gcd(1, 2)) + \operatorname{lcm}(3, \gcd(1, 1)) = \operatorname{lcm}(1, 1) + \operatorname{lcm}(1, 2) + \operatorname{lcm}(1, 1) + \operatorname{lcm}(2, 1) + \operatorname{lcm}(2, 1) + \operatorname{lcm}(3, 1) = 1 + 2 + 1 + 2 + 2 + 3 = 11$
In the second example, $\operatorname{lcm}(1, \gcd(3, 1)) + \operatorname{lcm}(1, \gcd(2, 2)) + \operatorname{lcm}(1, \gcd(1, 3)) + \operatorname{lcm}(2, \gcd(2, 1)) + \operatorname{lcm}(2, \gcd(1, 2)) + \operatorname{lcm}(3, \gcd(1, 1)) = \operatorname{lcm}(1, 1) + \operatorname{lcm}(1, 2) + \operatorname{lcm}(1, 1) + \operatorname{lcm}(2, 1) + \operatorname{lcm}(2, 1) + \operatorname{lcm}(3, 1) = 1 + 2 + 1 + 2 + 2 + 3 = 11$