BZOJ4804: 歐拉心算(莫比烏斯反演 線性篩)

題意

求$$\sum_1^n \sum_1^n \phi(gcd(i, j))$$

$T \leqslant 5000, N \leqslant 10^7$

Sol

延用BZOJ4407的作法

化到最後能夠獲得

$$\sum_{T = 1}^n \frac{n}{T} \frac{n}{T} \sum_{d \mid T}^n \phi(d) \mu(\frac{T}{d})$$

後面的那個是積性函數，直接篩出來

注意這個函數比較特殊，篩的時候須要分幾種狀況討論

1. $H(p) = p - 2$

2. $H(p^2) = p^2 - 2p + 1$

3. $H(p^{k + 1}) = H(p^k) * p$

// luogu-judger-enable-o2
#include<cstdio>
#include<algorithm>
#define LL long long 
using namespace std;
const int MAXN = 1e7 + 10, mod = 1e9 + 7;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int prime[MAXN], vis[MAXN], tot;
LL H[MAXN], low[MAXN];
void GetH(int N) {
    H[1] = vis[1] = 1;
    for(int i = 2; i <= N; i++) {
        if(!vis[i]) prime[++tot] = i, H[i] = i - 2, low[i] = i;
        for(int j = 1; j <= tot && i * prime[j] <= N; j++) {
            vis[i * prime[j]] = 1;
            if(!(i % prime[j])) {
                low[i * prime[j]] = low[i] * prime[j];
                if(low[i] == i) {
                    if(low[i] == prime[j]) H[i * prime[j]] = (H[i] * prime[j] + 1);
                    else H[i * prime[j]] = H[i] * prime[j];
                } 
                else H[i * prime[j]] = H[i / low[i]] * H[low[i] * prime[j]];
                break;
            }
            H[i * prime[j]] = H[i] * H[prime[j]];
            low[i * prime[j]] = prime[j];
        }
    }
    for(int i = 2; i <= N; i++) 
        H[i] = H[i - 1] + H[i];
}
int main() {
    GetH(1e7 + 5);
    int T = read();
    while(T--) {
        int N = read(), last;
        LL ans = 0;
        for(int i = 1; i <= N; i = last + 1) {
            last = N / (N / i);
            ans = ans + 1ll * (N / i) * (N / i) * (H[last] - H[i - 1]);
        } 
        printf("%lld\n", ans);
    }
    return 0;
}
/*
3
7001
123000
10000000
*/