hdu 6588 Function （数论+莫比乌斯反演）

 

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首先考虑分块，按照  的大小分块得到 

设 ，化简得到      （化简过程无法证明，也看不懂题解，我太菜了）

所以原式可以化成

右边直接枚举因子，可以在  求出来，所以我们考虑化简左边的式子，枚举  得到 

用常见的整除分块，预处理  前缀和，就可以在  求出来

所以原式 

整体复杂度 

#include 
#define ll long long //T了就换int 试试
#define int128 __int128
#define sc scanf
#define pr printf
using namespace std;
const int MAXN = 1e7 + 5;//用板子前先改范围

bool check[MAXN + 10];//值为 false 表示素数，值为 true 表示非素数
int phi[MAXN + 10];//欧拉函数表
int prime[MAXN + 10];//连续素数表
//int mu[MAXN + 10];//莫比乌斯函数
int tot;//素数的个数（从0开始
ll sub[MAXN + 10];
ll subi[MAXN + 10];
const ll mod = 998244353;
const ll inv2 = (mod + 1) / 2;
const ll inv6 = (mod + 1) / 6;
void jzk()
{
    //memset(check, false, sizeof(check));
    phi[1] = 1;
    //mu[1] = 1;
    check[1] = true;
    tot = 0;
    for (int i = 2; i < MAXN; i++)
    {
        if (!check[i])
        {
            prime[tot++] = i;
            phi[i] = i - 1;
            //mu[i] = -1;
        }
        for (int j = 0; j < tot; j++)
        {
            if (i * prime[j] >= MAXN)
                break;
            check[i * prime[j]] = true;
            if (i % prime[j] == 0)
            {
                phi[i * prime[j]] = phi[i] * prime[j];
                //mu[i * prime[j]] = 0;
                break;
            }
            else
            {
                phi[i * prime[j]] = phi[i] * (prime[j] - 1);
                //mu[i * prime[j]] = -mu[i];
            }
        }
    }
    for (int i = 1; i < MAXN; i++)
    {
        sub[i] = (sub[i - 1] + phi[i]) % mod;
        subi[i] = (subi[i - 1] + (ll)phi[i] * i) % mod;
    }
}
ll ck(int128 n)
{
    int128 l = 1, r = /*1e4;*/1e7;
    while (l + 1 < r)
    {
        int128 k = (l + r) / 2;
        if (k * k * k > n)
            r = k;
        else
            l = k;
    }
    if (r * r * r <= n)
        l = r;
    return (ll)l;
}
ll calc2(ll n)
{
    return n * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod;
}
ll calc1(ll n)
{
    return n * (n + 1) % mod * inv2 % mod;
}
ll calc0(ll n)
{
    return n;
}
inline int128 read()
{
    int128 X = 0, w = 0; char ch = 0;
    while (!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline void print(int128 x)
{
    if (x < 0) { putchar('-'); x = -x; }
    if (x > 9) print(x / 10);
    putchar(x % 10 + '0');
}
int main()
{
    //freopen("test1.in","r",stdin);
    jzk();
    int T;
    sc("%d", &T);
    while (T--)
    {
        int128 n;
        n = read();
        //sc("%lld", &n);
        int128 ans = 0;
        ll n3 = ck(n);
        int128 n33 = (int128)n3 * n3 * n3;
        ll j;
        for (ll T = 1; T <= n3 - 1; T = j + 1)
        {
            /*ll res = (3 * T * calc2((n3 - 1) / T) % mod + 3 * calc1((n3 - 1) / T) % mod + calc0((n3 - 1) / T)) % mod;
            res = res * phi[T];*/
            j = (n3 - 1) / ((n3 - 1) / T);
            ll res = ((subi[j] - subi[T - 1] + mod) * 3 * calc2((n3 - 1) / T) % mod + (sub[j] - sub[T - 1] + mod) * 3 * calc1((n3 - 1) / T) % mod + (sub[j] - sub[T - 1] + mod) * calc0((n3 - 1) / T) % mod) % mod;
            ans = (ans + res) % mod;
        }
        for (ll T = 1; T * T <= n3; T++)
        {
            if (n3 % T == 0)
            {
                ans = (ans + phi[T] * ((n / T) - ((n33 - 1) / T) + mod)) % mod;
                if (T * T != n3)
                    ans = (ans + phi[n3 / T] * ((n / (n3 / T)) - ((n33 - 1) / (n3 / T)) + mod)) % mod;
            }
        }
        print(ans);
        putchar('\n');
        //pr("%lld\n", ans);
    }
}