HDU-6390 GuGuFishtion(莫比乌斯反演)

HDU-6390 GuGuFishtion

求逆元时线性递推即可。

#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof(a))
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
#define sc scanf
#define itn int
#define STR clock_t startTime = clock();
#define END clock_t endTime = clock();cout << double(endTime - startTime) / CLOCKS_PER_SEC *1000<< "ms" << endl;
using namespace std;
const int N=1e6+5;
const long long mod=1e9+7;
const long long mod2=998244353;
const int oo=0x7fffffff;
const int sup=0x80000000;
typedef long long ll;
typedef unsigned long long ull;
template <typename it>void db(it *begin,it *end){while(begin!=end)cout<<(*begin++)<<" ";puts("");}
template <typename it>
string to_str(it n){string s="";while(n)s+=n%10+'0',n/=10;reverse(s.begin(),s.end());return s;}
template <typename it>int o(it a){cout<<a<<endl;return 0;}
inline ll mul(ll a,ll b,ll c){ll ans=0;for(;b;b>>=1,a=(a+a)%c)if(b&1)ans=(ans+a)%c;return ans;}
inline ll ksm(ll a,ll b,ll c){ll ans=1;for(;b;b>>=1,a=mul(a,a,c))if(b&1)ans=mul(ans,a,c);return ans;}
inline void exgcd(ll a,ll b,ll &x,ll &y){if(!b)x=1,y=0;else exgcd(b,a%b,y,x),y-=x*(a/b);}
int prime[N],tot=0;
short int mu[N];
bool vis[N]={0};
int phi[N];
ll f[N],inv[N];
void pre(){
    mu[1]=phi[1]=1;
    for(int i=2;i<N;i++){
        if(!vis[i])prime[++tot]=i,mu[i]=-1,phi[i]=i-1;
        for(int j=1;j<=tot&&i*prime[j]<N;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0){
                mu[i*prime[j]]=0;
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }else phi[i*prime[j]]=phi[i]*phi[prime[j]],mu[i*prime[j]]=-mu[i];
        }
    }
}
int m,n,p;
void fun(ll &x){
    if(x>=p)x-=x/p*p;
    if(x<0)x=x%p+p;
}
int main(){
   // STR IN
    pre();
    int t;cin>>t;
    while(t--){
        sc("%d%d%d",&m,&n,&p);
        if(n>m)swap(n,m);
        for(int i=1;i<=n;i++)f[i]=inv[i]=0;
        inv[1]=1;
        for(int i=2;i<=n;i++)inv[i]=1LL*(p-p/i)*inv[p%i]%p;
        for(int i=1;i<=n;i++){
            for(int j=i;j<=n;j+=i){
                ll x=1LL*i*mu[j/i]*inv[phi[i]];
                fun(x);
                f[j]+=x;
                fun(f[j]);

            }
        }
        ll ans=0;
        for(int i=1;i<=n;i++){
            ll x=1LL*(n/i)*(m/i)%p*f[i];
            fun(x);
            ans+=x;
            fun(ans);
        }
       printf("%lld\n",ans);
    }//END
}