2018沈阳网络赛 Convex Hull(莫比乌斯,容斥)

$Description$
图片说明

$Solution$
$gay(n)=f(n)=n^2\cdot \mu(n)^2$
$\sum_{i=1}^{n}\sum_{j=1}^{i}f(i)=(n+1)\sum_{i=1}^{n}f(i)-\sum_{i=1}^{n}i\cdot f(i)$
$\sum_{i=1}^{n}h(i)\cdot \mu(i)^2 \quad 考虑容斥,将i^2的倍数筛去.$
$算上他们的贡献就是h(i^2)\sum_{j=1}^{\frac{n}{i^2}}h(j).$
$ans=\sum_{i=1}^{\sqrt n}\mu(i)\cdot h(i^2)\sum_{j=1}^{\frac{n}{i^2}}h(j).$
$\sum_{i=1}^{n}h(i)\mu(i)^2 \\ =\sum_{i=1}^{n}h(i)\sum_{d^2|n}\mu(d) \\=\sum_{d=1}^{\sqrt n}\mu(d)\sum_{i=1}^{n}h(i)[d^2|i] \\=\sum_{d=1}^{\sqrt n}\mu(d)\sum_{i=1}^{\frac{n}{d^2}}h(i\cdot d^2) \\ 因为这里h是完全积性函数所以可以提出来\sum_{d=1}^{\sqrt n}\mu(d)\cdot h(d^2)\sum_{i=1}^{\frac{n}{d^2}}h(i)$

$Code$

#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof(a))
#define sc scanf
#define itn int
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
#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;
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 x,ll y,ll c){return (x*y-(ll)((long double)x/c*y)*c+c)%c;}
ll ksm(ll a,ll b,ll c){ll ans=1;for(;b;b>>=1,a=a*a%c)if(b&1)ans=ans*a%c;return ans;}
void exgcd(ll a,ll b,ll &x,ll &y){if(!b)x=1,y=0;else exgcd(b,a%b,y,x),y-=(a/b)*x;}
ll n,p;
ll pow_2(ll x){
    ll ans=mul(x,x+1,6LL*p);
    ans=mul(ans,2LL*x+1,6LL*p);
    ans/=6;
    return ans;
}
ll pow_3(ll x){
    ll ans=mul(x,x+1,2LL*p);
    ans/=2;
    ans=mul(ans,ans,p);
    return ans;
}
int prime[N],tot=0;
bool vis[N];
short int mu[N];
void pre(){
    mu[1]=1;
    for(int i=2;i<N;i++){
        if(!vis[i])prime[++tot]=i,mu[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;
                break;
            }else mu[i*prime[j]]=-mu[i];
        }
    }
}
int main(){
    pre();
    while(sc("%lld%lld",&n,&p)>0){
        ll ans=0,res=0;
        ll T=sqrt(n+0.5);
        for(ll i=1;i<=T;i++){
            ll x=mul(i*i,i*i,p);
            ans+=mu[i]*mul(pow_2(n/(i*i)),x,p);
            if(ans>=p)ans-=ans/p*p;
            if(ans<0)ans=(ans%p+p)%p;
            x=mul(x,i*i,p);
            res+=mu[i]*mul(pow_3(n/(i*i)),x,p);
            if(res>=p)res-=res/p*p;
            if(res<0)res=(res%p+p)%p;
        }
        printf("%lld\n",(mul(n+1,ans,p)-res+p)%p);
    }
}