BZOJ4916 神犇和蒟蒻

题意：求miu[i*i]和phi[i*i]的前缀和。n <= 1e9

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解：杜教筛。

第一问是搞笑的，答案是1。

第二问，发现phi[i*i] = phi[i] * i

这里杜教筛用的g(n) = n，h(n) = n²。

然后套路一波就没了。注意预处理的时候也要开long long

#include <cstdio>
#include <cstring>

typedef long long LL;
const int N = 1100010, T = 1100008;
const LL MO = 1e9 + 7, inv6 = 166666668;

int p[N], top, phi[N];
LL F[N], F2[N], n, inv2;
bool vis[N];

inline LL H(LL x) {
    x %= MO;
    return (x << 1 | 1) % MO * x % MO * (x + 1) % MO * inv6 % MO;
}

inline void getp(int n) {
    phi[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(!vis[i]) {
            p[++top] = i;
            phi[i] = i - 1;
        }
        for(int j = 1; j <= top && i * p[j] <= n; j++) {
            vis[i * p[j]] = 1;
            if(i % p[j] == 0) {
                phi[i * p[j]] = phi[i] * p[j];
                break;
            }
            phi[i * p[j]] = phi[i] * (p[j] - 1);
        }
    }
    for(int i = 1; i <= n; i++) {
        F[i] = (F[i - 1] + 1ll * i * phi[i] % MO) % MO;
    }
    return;
}

LL solve(LL x) {
    //printf("solve %lld 
", x);
    if(x <= 0) return 0;
    if(x <= T) return F[x];
    if(F2[n / x] != -1) return F2[n / x];
    //printf("solve %lld 
", x);
    LL ans = H(x);
    for(LL i = 2, j; i <= x; i = j + 1) {
        j = x / (x / i);
        ans -= (i + j) % MO * ((j - i + 1) % MO) % MO * inv2 % MO * solve(x / i) % MO;
        ans = (ans % MO + MO) % MO;
    }
    return F2[n / x] = ans;
}

int main() {
    inv2 = (MO + 1) / 2;
    memset(F2, -1, sizeof(F2));
    getp(T);
    scanf("%lld", &n);
    printf("1
%lld
", solve(n));
    return 0;
}