E - GuGuFishtion HDU - 6390 （欧拉函数，反演）

Today XianYu is too busy with his homework, but the boring GuGu is still disturbing him!!!!!!
At the break time, an evil idea arises in XianYu's mind.
‘Come on, you xxxxxxx little guy.’
‘I will give you a function

ϕ (x)

G u (a, b) = ϕ ( a b ) ϕ ( a ) ϕ ( b )

ϕ (x)

(\sum a = 1 m \sum b = 1 n G u (a, b))

ϕ (x)

InputInput contains an integer $T$ OutputPlease output exactly $T$ Sample Input

1
5 7 23

Sample Output

2

Solution：

phi(ab) = phi(a)*phi(b)*gcd(a,b)/phi(gcd(a,b))
然后在去枚举gcd就行了
递推逆元爆int？

Code：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+100;
typedef long long ll;
int fac[maxn],mu[maxn],phi[maxn];
int g[maxn];
ll inv[maxn];
void init()
{
    for(int i=1; i<maxn; ++i)
        fac[i]=i;
    phi[1]=mu[1]=1;
    for(int i=2; i<maxn; ++i)
    {
        if(fac[i]==i)
            for(int j=i<<1; j<maxn; j+=i)
                fac[j]=i;
        if(i/fac[i]%fac[i])
            phi[i]=(fac[i]-1)*phi[i/fac[i]],mu[i]=-mu[i/fac[i]];
        else
            phi[i]=fac[i]*phi[i/fac[i]],mu[i]=0;
    }
}
int n,m,p;

signed main()
{
    ios::sync_with_stdio(0);
    init();
    int T;
    cin>>T;
    while(T--)
    {
        cin>>n>>m>>p;
        int M=min(n,m);

        for(int i=1; i<=M; i++)
        {
            g[i]=0;
            int MM=min(n/i,m/i);
            int A=n/i;
            int B=m/i;
            for(int d=1; d<=MM; d++)
                g[i] += 1ll*mu[d]*(n/i/d)*(m/i/d)%p,g[i]%=p;
            if(g[i]<0)
                g[i] += p;
        }

        int ans=0;
        inv[1]=1;
        for( int i=2; i<=M; i++)
        {
            inv[i]=1ll*(p-p/i)*inv[p%i]%p;
        }
        for(int k=1; k<=M; k++)
        {
            ans=(1ll*ans+1ll*k*inv[phi[k]]%p*1ll*g[k]%p)%p;
        }
        cout<<ans<<endl;


    }
// 5 1000000 1000000 1000000007
}

原文地址：https://www.cnblogs.com/zhangbuang/p/11881187.html