BZOJ4916: 神犇和蒟蒻(杜教筛)

题意

求

$$sum_{i = 1}^n mu(i^2)$$

$$sum_{i = 1}^n phi(i^2)$$

$n leqslant 10^9$

Sol

zz的我看第一问看了10min。

感觉自己智商被侮辱了qwq

基础太垃圾qwq。

算了正经点吧，第一问答案肯定是$1$，还不明白的重学反演吧。

第二问其实也不难

定理：

$phi(i^2) = iphi(i)$

$sum_{d | n} phi(d) = n$

显然$i$

考虑杜教筛的套路式子

$$g(1)s(n) = sum_{i = 1}^n g(i)s(frac{n}{i}) - sum_{i = 2}^n g(i)s(frac{n}{i})$$

当我们选择$g(i) = id(i) = i$时卷积的前缀和是比较好算的

$(g * s)(i) = sum_{i = 1}^n i^2 = frac{n * (n + 1) * (2n + 1)}{6}$

然后上杜教筛就行了

$$s(n) = frac{n * (n + 1) * (2n + 1)}{6} - sum_{i = 2}^n i phi(frac{n}{i})$$

人傻自带大常数

#include<cstdio>
#include<map>
#define LL long long 
using namespace std;
const int MAXN = 1e7 + 10, mod = 1e9 + 7;
const LL inv = 166666668;
int N, prime[MAXN], vis[MAXN], tot;
LL phi[MAXN];
map<int, LL> ans;
void GetPhi(int N) {
    vis[1] = phi[1] = 1;
    for(int i = 2; i <= N; i++) {
        if(!vis[i]) prime[++tot] = i, phi[i] = i - 1;
        for(int j = 1; j <= tot && i * prime[j] <= N; j++) {
            vis[i * prime[j]] = 1;
            if(!(i % prime[j])) {phi[i * prime[j]] = phi[i] * prime[j]; break;}
            phi[i * prime[j]] = phi[i] * phi[prime[j]];
        }
    }
    for(int i = 1; i <= N; i++) phi[i] = (1ll * i * phi[i] % mod +  phi[i - 1] % mod) % mod;
}
LL Query(LL x) {
    return (x * (x + 1) / 2) % mod;
}
LL S(LL N) {
    if(ans[N]) return ans[N];
    if(N <= 1e7) return phi[N];
    LL sum = N * (N + 1) % mod * (2 * N + 1) % mod * inv % mod, last = 0;
    for(int i = 2; i <= N; i = last + 1) {
        last = N / (N / i);
        sum -= S(N / i) % mod * (Query(last) - Query(i - 1)) % mod;
        sum = (sum + mod) % mod;
    }
    return ans[N] = (sum % mod + mod) % mod;
}
int main() {
    GetPhi(1e7);
    scanf("%d", &N);
    printf("1
%lld", S(N));
    return 0;
}