P1447 [NOI2010]能量采集（mobius反演）

P1447 [NOI2010]能量采集

式子化简

显然题目就是要我们求 $\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} 2gcd(i, j) - 1$

$= 2\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} gcd(i, j) - nm$

转化为我们要求 $\sum_{i = 1} ^{n} \sum_{j = 1} ^{m} gcd(i, j)$

$= \sum_{d = 1} ^{n}d\sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{m}{d}} gcd(i, j) == 1$

套上 $mobius$

$= \sum_{d = 1} ^{n}d\sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{m}{d}} \sum_{k \mid gcd(i, j)} \mu(k)$

$= \sum_{d = 1} ^{n}d\sum_{i = 1} ^{\frac{n}{d}} \sum_{j = 1} ^{\frac{m}{d}} \sum_{k \mid gcd(i, j)} \mu(k)$

$= \sum_{d = 1} ^{n} d\sum_{k = 1} ^{\frac{n}{d}}\mu(k) \lfloor\frac{n}{dk}\rfloor \lfloor\frac{m}{dk}\rfloor$

另 $t = dk$

$= \sum_{t = 1} ^{n} \lfloor\frac{n}{t}\rfloor \lfloor\frac{m}{t}\rfloor \sum_{d \mid t}d\mu(\frac{t}{d})$

$mobius$反演有 $\sum_{d\mid n}\frac{\mu(d)}{d} = \frac{\phi(n)}{n}$

$= \sum_{t = 1} ^{n} \lfloor\frac{n}{t}\rfloor \lfloor\frac{m}{t}\rfloor \phi(t)$

代码

/*
  Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>

#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;

inline ll read() {
    ll f = 1, x = 0;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-')    f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f * x;
}

const int N = 1e7 + 10;

bool st[N];

vector<int> prime;

int n, m;

ll phi[N];

void mobius() {
    st[0] = st[1] = phi[1] = 1;
    for(int i = 2; i < N; i++) {
        if(!st[i]) {
            prime.pb(i);
            phi[i] = i - 1;
        }
        for(int j = 0; j < prime.size() && i * prime[j] < N; j++) {
            st[i * prime[j]] = 1;
            if(i % prime[j] == 0) {
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
            phi[i * prime[j]] = phi[i] * (prime[j] - 1);
        }
    }
    for(int i = 1; i < N; i++) phi[i] += phi[i - 1];
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    mobius();
    ll n = read(), m = read();
    if(n > m) swap(n, m);
    ll ans = 0;
    for(ll l = 1, r; l <= n; l = r + 1) {
        r = min(n / (n / l), m / (m / l));
        ans += (n / l) * (m / l) * (phi[r] - phi[l - 1]);
    }
    printf("%lld\n", 2 * ans - n * m);
	return 0;
}