Description

感覺好迷茫啊，很多變換看的一臉懵逼卻又不知道去哪裏學。一道題做一上午也是沒誰了，，

首先按照套路反演化到最後應該是這個式子

$$ans = \sum_{d = 1}^n d^k \sum_{i = 1}^{\frac{n}{d}} \frac{n}{di} \frac{m}{di} \mu(i)$$

這樣就可以$O(n)$計算

繼續往下推，考慮$\frac{n}{di} \frac{m}{di}$對答案的貢獻

設$T = id$

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

後面那一坨是狄利克雷卷積的形式，顯然是積性函數，可以直接篩

然後我在這裏懵了一個小時，，

設$H(T) = \sum_{d \mid T} ^ T d^k \mu(\frac{T}{d})$

那麽當$T = p^a$式，上面的式子中只有$\frac{T}{d} = 1$或$\frac{T}{d} = p$式，$\mu(\frac{T}{d})$才不為$0$

那麽把式子展開$H(p^{a + 1}) = H(p^a) * (p^k)$

// luogu-judger-enable-o2
#include<cstdio>
#include<algorithm>
#define
 
 LL long long 
using namespace std;
const int MAXN = 5 * 1e6 + 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 T, K;
int prime[MAXN], vis[MAXN], tot, mu[MAXN];
LL H[MAXN], low[MAXN];
LL fastpow(LL a, LL p) {
    LL base = 1;
    while(p) {
        if(p & 1) base = (base * a) % mod;
        a = (a * a) % mod; p >>= 1;
    }
    return base;
} 
void GetH(int N) {
    vis[1] = H[1] = mu[1] = low[1] = 1;
    for(int i = 2; i <= N; i++) {
        if(!vis[i]) prime[++tot] = i, mu[i] = -1, H[i] = (-1 + fastpow(i, K) + mod) % mod, low[i] = i;
        for(int j = 1; j <= tot && i * prime[j] <= N; j++) {
            vis[i * prime[j]] = 1;
            if(!(i % prime[j])) {
                mu[i * prime[j]] = 0; low[i * prime[j]] = (low[i] * prime[j]) % mod; 
                if(low[i] == i) 
                    //H[i * prime[j]] = (H[i] + fastpow((i * prime[j]), K)) % mod;
                    H[i * prime[j]] = H[i] * (fastpow(prime[j], K)) % mod;
                else H[i * prime[j]] = H[i / low[i]] * H[prime[j] * low[i]] % mod;
                break;
            } 
            mu[i * prime[j]] = mu[i] * mu[prime[j]] % mod;
            H[i * prime[j]] = H[i] * H[prime[j]] % mod;
            low[i * prime[j]] = prime[j] % mod;
        }
    }
    for(int i = 2; i <= N; i++) H[i] = (H[i] + H[i - 1] + mod) % mod;
}
int main() {
    T = read(); K = read();
    GetH(5000001);
    while(T--) {
        int N = read(), M = read(), last;
        LL ans = 0;
        if(N > M) swap(N, M);
        for(int T = 1; T <= N; T = last + 1) {
            last = min(N / (N / T), M / (M / T));
            ans = (ans + (1ll * (N / T) * (M / T) % mod) * (H[last] - H[T - 1] + mod)) % mod;
        }
        printf("%lld\n", ans % mod);
    }
    return 0;
}
/*
2 5000000
7 8
123 456
4999999 5000000
*/

BZOJ4407: 於神之怒加強版(莫比烏斯反演 線性篩)