hdu5528Count a * b(数论)

题目链接https://cn.vjudge.net/problem/HDU-5528
Marry likes to count the number of ways to choose two non-negative integers a and b less than m to make a×b mod m≠0.

Let's denote f(m)as the number of ways to choose two non-negative integers a and b less than m to make a×b mod m≠0.

She has calculated a lot of f(m) for different mm, and now she is interested in another function g(n)=∑m|nf(m). For example, g(6)=f(1)+f(2)+f(3)+f(6)=0+1+4+21=26. She needs you to double check the answer.



Give you nn. Your task is to find g(n) modulo 2^64.
Input
The first line contains an integer T indicating the total number of test cases. Each test case is a line with a positive integer nn.

1≤T≤20000
1≤n≤10^9
Output
【hdu5528Count a * b(数论)】For each test case, print one integer ss, representing g(n) modulo 2^64.
Sample Input

2 6 514

Sample Output
26 328194

听说结果没有大于等于2^64的,直接用long long算???
推导见大神博客https://blog.csdn.net/Coldfresh/article/details/82189233

#include #include #include #include #include #include #include #include #include #include #include #include using namespace std; typedef long long ll; const int N=4e4+5; bool p[N]; int prime[N],cnt=0; void init(){ p[0]=p[1]=true; for(int i=2; i


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