【HDU 5528】Count a * b（推导）

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Problem Description
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 m, and now she is interested in another function $g(n)=\sum_{m|n}f(m)$. 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 $n$. 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 n.

$1≤T≤20000$
$1≤n≤10^9$

Output
For each test case, print one integer s, representing g(n) modulo 264.

Sample Input
2
6
514

Sample Output
26
328194

Source
2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

先推一下$f$:

$$f(m)=m^2-\sum_{i=1}^m\sum_{j=1}^mm|i*j\\= m^2-\sum_{i=1}^m\sum_{j=1}^m\frac{m}{(m,i)}|\frac{i}{(m,i)}*j$$
可以发现 $\frac{m}{(m,i)}和\frac{i}{(m,i)}$已经是互质了，所以要整除m， $j$必须是 $\frac{m}{(m,i)}$的倍数，其数量显然是 $(m,i)$个，则： $$f(m)=m^2-\sum_{i=1}^m(m,i)$$
这里是老套路，去枚举gcd：最终就有： $$f(m)=m^2-\sum_{d|m}d*\phi(\frac{m}{d})$$

来看$g$,所以：

$$g(n)=\sum_{m|n}m^2-\sum_{m|n}\sum_{d|m}d\phi(\frac{m}{d})$$
把d拿到最外层： $$g(n)=\sum_{m|n}m^2-\sum_{d|n}d\sum_{d|m,m|n}\phi(\frac{m}{d})\\= g(n)=\sum_{m|n}m^2-\sum_{d|n}d\sum_{k|\frac{n}{d}}\phi(k)\\= g(n)=\sum_{m|n}m^2-\sum_{d|n}n$$

到这里就是去枚举n的所以因子了，用唯一分解来做，然后搜索枚举因子即可。
代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define ull  unsigned long long
#define ll long long
#define ul unsigned int
#define maxn 40000
using namespace std;
bool isP[maxn];
int prime[maxn];
int cnt;
void init()
{
    for(int i=2;i<maxn;i++)
    {
        if(!isP[i])prime[cnt++]=i;
        for(int j=0;j<cnt&&i*prime[j]<maxn;j++)
        {
            isP[i*prime[j]]=true;
            if(i%prime[j]==0)break;
        }
    }
}
int fac[50];
int e[50];
int k;
void getFac(int x)
{
    k=0;
    for(int i=0;prime[i]*prime[i]<=x;i++)
    {
        if(x%prime[i]==0)
        {
            fac[k]=prime[i];
            e[k]=0;
            while(x%prime[i]==0)
            {
                x/=prime[i];
                ++e[k];
            }
            ++k;
        }
    }
    if(x>1){fac[k]=x;e[k++]=1;}
}
ull ans=0;
void dfs(int cur,ull temp)
{
    if(cur==k)
    {
        ans+=temp*temp;
        return;
    }
    dfs(cur+1,temp);
    ull now=temp;
    for(int i=1;i<=e[cur];i++)
    {
        now*=fac[cur];
        dfs(cur+1,now);
    }
}
ull mod=0;
int main()
{
    mod=~mod;
    init();
    int t;
    cin>>t;
    int n;
    while(t--)
    {
        scanf("%d",&n);
        int total=1;
        getFac(n);
        ans=0;
        dfs(0,1);
        for(int i=0;i<k;i++)total*=e[i]+1;
        ull _n=n;
        _n*=total;
        _n=mod-_n+1;
        cout<<ans+_n<<endl;
    }
    return 0;
}