2018 ACM-ICPC 沈阳网络赛C Convex hull

题目链接：https://nanti.jisuanke.com/t/31444

考虑杜教筛就跑偏了…
定义 gay(i)=i2∗μ2(i) g a y ( i ) = i 2 ∗ μ 2 ( i )
∑ni=1∑ij=1gay(i)=∑ni=1(n−i+1)gay(i)=∑ni=1(n−i+1)i2μ2(i) ∑ i = 1 n ∑ j = 1 i g a y ( i ) = ∑ i = 1 n ( n − i + 1 ) g a y ( i ) = ∑ i = 1 n ( n − i + 1 ) i 2 μ 2 ( i )
有 μ2(n)=∑i2|nμ(i) μ 2 ( n ) = ∑ i 2 | n μ ( i )
因此上式 =∑ni=1(n−i+1)i2(∑j2|iμ(j)) = ∑ i = 1 n ( n − i + 1 ) i 2 ( ∑ j 2 | i μ ( j ) )
枚举因子变成枚举倍数
∑ni=1(n−i+1)i2(∑j2|iμ(j))=∑n√i=1μ(i)(∑⌊ni2⌋j=1(n−i2j+1)i4j2)=∑n√i=1μ(i)(∑⌊ni2⌋j=1(i4j2(n+1)−i6j3)) ∑ i = 1 n ( n − i + 1 ) i 2 ( ∑ j 2 | i μ ( j ) ) = ∑ i = 1 n μ ( i ) ( ∑ j = 1 ⌊ n i 2 ⌋ ( n − i 2 j + 1 ) i 4 j 2 ) = ∑ i = 1 n μ ( i ) ( ∑ j = 1 ⌊ n i 2 ⌋ ( i 4 j 2 ( n + 1 ) − i 6 j 3 ) )
∑ni=1i2=n∗(n+1)∗(2n+1)6,∑ni=1i3=n2(n+1)24 ∑ i = 1 n i 2 = n ∗ ( n + 1 ) ∗ ( 2 n + 1 ) 6 , ∑ i = 1 n i 3 = n 2 ( n + 1 ) 2 4
直接开写吧，注意过程爆LL。
∑n√i=1μ(i)(∑⌊ni2⌋j=1(i4j2(n+1)−i6j3)) ∑ i = 1 n μ ( i ) ( ∑ j = 1 ⌊ n i 2 ⌋ ( i 4 j 2 ( n + 1 ) − i 6 j 3 ) )
=∑n√i=1μ(i)(∑⌊ni2⌋j=1(i4j2(n+1)−∑⌊ni2⌋j=1i6j3)) = ∑ i = 1 n μ ( i ) ( ∑ j = 1 ⌊ n i 2 ⌋ ( i 4 j 2 ( n + 1 ) − ∑ j = 1 ⌊ n i 2 ⌋ i 6 j 3 ) )

#define others
#ifdef poj
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#endif // poj
#ifdef others
#include 
#include 
#include 
#endif // others
//#define file
#define all(x) x.begin(), x.end()
using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
#define eps 1e-8
const double pi = acos(-1.0);

typedef long long LL;
typedef __int128 DLL;
typedef unsigned long long ULL;
void umax(LL &a, LL b) {
    a = max(a, b);
}
void umin(LL &a, LL b) {
    a = min(a, b);
}
int dcmp(double x) {
    return fabs(x) <= eps?0:(x > 0?1:-1);
}
void file() {
    freopen("data_in.txt", "r", stdin);
    freopen("data_out.txt", "w", stdout);
}

DLL mod = 1e9+7;

DLL Pow(DLL a,DLL b) {
    DLL res=1;
    a%=mod;
    for(; b; b>>=1) {
        if(b&1)res=res*a%mod;
        a=a*a%mod;
    }
    return res;
}
//
//void print(DLL x) {
//    if(x < 0) {
//        x = -x;
//        putchar('-');
//    }
//    if(x > 9) print(x/10);
//    putchar(x%10 + '0');
//}
//#define iostart

#define pb(x) push_back(x)
namespace solver {
    const int maxn = 1100000;
    int cnt;
    int p[maxn/10];
    int mob[maxn];
    bool tag[maxn];
    void shai() {
        cnt = 0;
        mob[1] = 1;
        for (int i = 2; i < maxn; i++) {
            if (!tag[i]) {
                mob[i] = -1;
                p[cnt++] = i;
            }
            for (int j = 0; j < cnt && i * p[j] < maxn; j++){
                tag[i*p[j]] = 1;
                if (i % p[j] == 0){
                    mob[i*p[j]] = 0;
                    break;
                }
                mob[i*p[j]] = -mob[i];
            }
        }
    }
    DLL get2(DLL n) {
        return (n*(n+1)*(2*n+1)/6)%mod;
    }
    DLL get3(DLL n) {
        DLL v = n * (n + 1) / 2;
        v %= mod;
        return v*v%mod;
    }
    void solve() {
        shai();
        LL n, p;
        while (~scanf("%lld%lld", &n, &p)) {
            mod = p;
            LL res = 0;
            for (DLL i = 1; i <= sqrt(n); i++){
                DLL i4 = Pow(i, 4);
                DLL i6 = Pow(i, 6);
                DLL v = n/(i*i);
                DLL temp = i4*(n+1)%mod*get2(v)%mod;
                temp -= i6*get3(v)%mod;
                temp %= mod;
                temp += mod;
                temp %= mod;
                res += (DLL)mob[i] * temp % mod;
                res %= mod;
                res += mod;
                res %= mod;
            }
            cout << res << '\n';
        }
    };
}

int main() {
#ifdef iostart
    ios::sync_with_stdio(0);
    cin.tie(0);
#endif // iostart
//    file();
    solver::solve();
    return 0;
}