2019 Multi-University Training Contest 1 - 1011 - Function - 数论

http://acm.hdu.edu.cn/showproblem.php?pid=6588

Find a new learned within the gcd of m and n and fast method for finding. That is below S1.

---

① requirements:

$ \sum\limits_{i=1}^{n}gcd(m,i) $

② enumeration d:

$ \sum\limits_{d|m} d \sum\limits_{i=1}^{n} [gcd(m,i)==d] $

Obviously ③:

$ \sum\limits_{d|m} d \sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor} [gcd(\frac{m}{d},i)==1] $

This step has been to seek a recursively, Kiki said to be \ (O (the n-^ {\ FRAC {3} {4}}) \) , but the solution to a problem can continue down.

④ In order to facilitate a direct consideration $ \ sum \ limits_ {i = 1} ^ {n} [gcd (m, i) == 1] $, inversion (approximate):

$ \sum\limits_{i=1}^{n} \sum\limits_{d|gcd(m,i)} \mu(d) $

⑤ swap the order of enumeration d, it is clear that the contribution will be in multiples of n d within one mu (d):

$ \sum\limits_{d|m} \mu(d) \lfloor\frac{n}{d}\rfloor $

---

The following is based on the realization of the solution to a problem, but a version of __int64.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef __int64 lll;

const int mod = 998244353;
const int MAXN = 10000000;

int phi[MAXN + 1];
int pri[MAXN + 1], pritop;
bool notpri[MAXN + 1];

void sieve() {
    int n = MAXN;
    pri[1] = phi[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(!pri[i])
            pri[++pritop] = i, phi[i] = i - 1;
        for(int j = 1, tmp; j <= pritop && (tmp = i * pri[j]) <= n; j++) {
            pri[tmp] = 1;
            if(i % pri[j])
                phi[tmp] = phi[i] * phi[pri[j]];
            else {
                phi[tmp] = phi[i] * pri[j];
                break;
            }
        }
    }
}

ll S1(lll n, int m) {
    //sigma gcd(i,m) [1,n]
    ll res = 0;
    for(int T = 1; T * T <= m; ++T) {
        if(!(m % T)) {
            res += (n / T) * phi[T];
            if(T * T != m) {
                res += (n / (m / T)) * phi[(m / T)];
            }
        }
    }
    res %= mod;
    return res;
}

ll qpow(ll x, int n) {
    ll res = 1;
    while(n) {
        if(n & 1)
            res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}

const int inv2 = qpow(2ll, mod - 2);
const int inv6 = qpow(6ll, mod - 2);

ll sigma1(ll x) {
    return x * (x + 1ll) % mod * inv2 % mod;
}

ll sigma2(ll x) {
    return x * (x + 1ll) % mod * (2ll * x + 1ll) % mod * inv6 % mod;
}

ll S2_1(int r, int T) {
    int c = r / T;
    ll res = 0;
    res += 3ll * T * sigma2(c);
    res += 3ll * sigma1(c);
    res += c;
    res %= mod;
    return res;
}

ll S2(int r) {
    ll res = 0;
    for(int T = 1; T <= r; ++T) {
        res += 1ll * phi[T] * S2_1(r, T) % mod;
    }
    res %= mod;
    return res;
}

ll S0(lll n) {
    lll i, i3;
    for(i = 1;; ++i) {
        lll tmp = i * i * i;
        if(tmp > n) {
            --i;
            break;
        } else
            i3 = tmp;
    }
    ll res = 0;
    res += S1(n, i) - S1(i3 - 1, i);
    res += S2(i - 1);
    res = (res % mod + mod) % mod;
    return res;
}

inline lll read() {
    lll x = 0;
    char c;
    do {
        c = getchar();
    } while(c < '0' || c > '9');
    do {
        x = (x << 3) + (x << 1) + c - '0';
        c = getchar();
    } while(c >= '0' && c <= '9');
    return x;
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    sieve();
    int T;
    cin >> T;
    lll n;
    while(T--) {
        n = read();
        cout << S0(n) << endl;
    }
}

In fact, specific ideas or first divided into two parts to count, but I did not count the S1 led to a T.