[ACM-ICPC 2018 沈阳网络赛] C. Convex Hull （莫比乌斯系数容斥）

We define a function gay(i)gay(i):

 

displaystyle gay(i)=left{ egin{array}{rcl} & 0, quad& if quad i = k*x*x, x > 1, k geq 1 \ & i * i ,quad& else end{array} ight.gay(i)={​0,i∗i,​ifi=k∗x∗x,x>1,k≥1else​

 

Now your task is to calculate

displaystyle sum_{num=1}^n (sum_{i=1}^{num} gay(i)) mod pnum=1∑n​(i=1∑num​gay(i))modp

 

Input

Multiple test cases. Please use EOF.

In each test case, there are two integers nn, pp, which are described above.

1 leq n leq 10^{10},1 leq p leq 10^{11}1≤n≤1010,1≤p≤1011.

The number of test cases is no more than 100100.

Output

For each test case print the answer in one line.

样例输入复制

1 10
8 19230817

样例输出复制

1
396

SOLUTION:
看到完全平方因子可以想到 莫比乌斯函数 ，当完全平方因子的时候，mu为0
这个题让求所有的不含完全平方因子的数，容斥一下，
对于 关于完全平方因子的容斥，枚举平方因子，其容斥系数为mu
这篇写的不错： https://blog.csdn.net/qkoqhh/article/details/82532516#commentsedit
CODE:

#include<bits/stdc++.h>

using namespace std;

typedef unsigned long long ll;

const ll maxn=1e5+10;

pair<__int128,__int128> p[maxn];

ll mod;
bool vis[maxn];
int prime[maxn];
int mu[maxn];

void init(){
    mu[1]=1;
    int tot=0;
    for(int i=2;i<maxn;i++){
        if(!vis[i]){
            prime[tot++]=i;
            mu[i]=-1;
        }
        for(int j=0;j<tot;j++){
            if(i*prime[j]>maxn) break;
            vis[i*prime[j]]=true;
            if(i%prime[j]==0){
                mu[i*prime[j]]=0;
                break;
            }else{
                mu[i*prime[j]]=-mu[i];
            }
        }
    }

}

ll pingfang(ll n){
    __int128 x=n,y=n+1,z=2*n+1;
    if(x%2==0) x/=2;
    else if(y%2==0) y/=2;
    else z/=2;
    if(x%3==0) x/=3;
    else if(y%3==0) y/=3;
    else z/=3;
    return (x%mod*y%mod*z%mod);
}

ll lifang(ll n){
    __int128 x=n,y=n+1;
    if(x%2==0) x/=2;
    else y/=2;
    return x%mod*x%mod*y%mod*y%mod;
}

int main(){
    ll n;
    init();
    while(scanf("%lld%lld",&n,&mod)!=EOF){
        for(ll i=0;i*i<=n;i++){
            p[i].first=i%mod*i%mod*i%mod*i%mod;
            p[i].second=i%mod*i%mod*i%mod*i%mod*i%mod*i%mod;
        }
        ll ans=0;
        for(ll i=1;i*i<=n;i++){
            ans+=mod+(n+1)%mod*p[i].first%mod*pingfang(n/(i*i))%mod*mu[i]-p[i].second%mod*lifang(n/(i*i))%mod*mu[i];
            ans=(ans+mod)%mod;
        }
        printf("%lld
",ans);
    }
}