求：
[sum_{i = 1}^{n} mu(i ^ 2) ~~~ sum_{i = 1}^{n} varphi(i ^ 2) \
n leq 1e9
]
第一问：我有个写法可惜这里空间太小写不下:
...
printf("1
");
...
下面我们来考虑第二问：
对于一个素因子(p)，我们有：
[varphi(p^k) = (p - 1) imes p ^{k - 1}\
varphi(p ^ {2k}) = (p - 1) imes p ^ {2k - 1}\
= (p - 1) imes p ^ {k - 1} imes p^{k}\
= varphi(p ^ k) * p ^ k
]
所以:
[f(n) = varphi(n ^ 2) \sum_{i = 1}^{n} varphi(i ^ 2) = sum_{i = 1}^{n} i imes varphi(i) \
]
构造$g(n) = n $, 有:
[(f * g)(n) = sum_{d | n} d imes varphi(d) imes frac{n}{d}\
= n sum_{d | n} varphi(d) = n ^ 2
]
有：$$1^2 + 2^2 + 3^2+ dots + n^2 = frac{n(n + 1)(2n + 1)}{6}$$
然后杜教筛搞一下就可以了。
Codes
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
#define rep(i, a, b) for(int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
#define drep(i, a, b) for(int i = (a), i##_end_ = (b); i >= i##_end_; --i)
#define clar(a, b) memset((a), (b), sizeof(a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
typedef long long LL;
typedef long double LD;
int read() {
char ch = getchar();
int x = 0, flag = 1;
for (;!isdigit(ch); ch = getchar()) if (ch == '-') flag *= -1;
for (;isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
return x * flag;
}
void write(int x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) write(x / 10);
putchar(x % 10 + 48);
}
const int Maxn = 3000009, Mod = (int)1e9 + 7;
__gnu_pbds :: cc_hash_table <int, LL> Sum;
LL isnprime[Maxn], prime[Maxn], tot;
LL phi[Maxn], prefix[Maxn];
void init() {
phi[1] = 1;
rep (i, 2, Maxn - 1) {
if (!isnprime[i]) {
prime[++tot] = i;
phi[i] = i - 1;
}
for (int j = 1, k; j < Maxn && (k = prime[j] * i) < Maxn; ++j) {
isnprime[k] = 1;
if (i % prime[j] == 0) {
phi[k] = prime[j] * phi[i];
break;
}else phi[k] = (prime[j] - 1) * phi[i];
}
}
rep (i, 1, Maxn - 1) prefix[i] = (prefix[i - 1] + phi[i] * 1ll * i % Mod) % Mod;
}
LL CalcPhiSquare(int val) {
if (val < Maxn) return prefix[val];
if (Sum[val]) return Sum[val];
LL res = val * (val + 1ll) % Mod * (val * 2ll + 1) % Mod * 166666668ll % Mod;
for (int l = 2, r; l <= val; l = r + 1) {
r = val / (val / l);
res -= ((r + 1ll) * r % Mod * 500000004ll % Mod - l * (l - 1ll) % Mod * 500000004ll % Mod + Mod) % Mod * CalcPhiSquare(val / l);
(res += Mod) %= Mod;
}
return Sum[val] = res;
}
void solve() {
int Index = read();
printf("1
%lld
", CalcPhiSquare(Index));
}
int main() {
freopen("BZOJ4916.in", "r", stdin);
freopen("BZOJ4916.out", "w", stdout);
init();
solve();
#ifdef Qrsikno
debug("
Running time: %.3lf(s)
", clock() * 1.0 / CLOCKS_PER_SEC);
#endif
return 0;
}