【HDU 5528】Count a * b（推导）

最新推荐文章于 2021-01-24 22:09:46 发布

coldfresh

最新推荐文章于 2021-01-24 22:09:46 发布

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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.
这里写图片描述
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}^{m} m | i * j = m^{2} - \sum_{i = 1}^{m} \sum_{j = 1}^{m} \frac{m}{(m, i)} | \frac{i}{(m, i)} * j

$f(m)=m^2-\sum_{i=1}^m\sum_{j=1}^mm|i*j\\= m^2-\sum_{i=1}^m\sum_{j=1}^m\frac{m}{(m,i)}|\frac{i}{(m,i)}*j$
可以发现

m(m,i)和i(m,i) m ( m , i ) 和 i ( m , i ) $\frac{m}{(m,i)}和\frac{i}{(m,i)}$ 已经是互质了，所以要整除m，

j j $j$ 必须是

\frac{m}{(m, i)}

$\frac{m}{(m,i)}$ 的倍数，其数量显然是

(m,i) ( m , i ) $(m,i)$ 个，则：

f (m) = m 2 - \sum i = 1 m (m, i)

$f(m)=m^2-\sum_{i=1}^m(m,i)$
这里是老套路，去枚举gcd：最终就有：

f (m) = m 2 - \sum d | m d * ϕ (m d)

$f(m)=m^2-\sum_{d|m}d*\phi(\frac{m}{d})$

来看 $g$ ,所以：

g (n) = \sum_{m | n} m^{2} - \sum_{m | n} \sum_{d | m} d ϕ (\frac{m}{d})

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

g (n) = \sum m | n m 2 - \sum d | n d \sum d | m, m | n ϕ (m d) = g (n) = \sum m | n m 2 - \sum d | n d \sum k | n d ϕ (k) = g (n) = \sum m | n m 2 - \sum d | n n

$g(n)=\sum_{m|n}m^2-\sum_{d|n}d\sum_{d|m,m|n}\phi(\frac{m}{d})\\= g(n)=\sum_{m|n}m^2-\sum_{d|n}d\sum_{k|\frac{n}{d}}\phi(k)\\= g(n)=\sum_{m|n}m^2-\sum_{d|n}n$

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;
}