传送门
拆开变成
[prod_{i=1}^{n}sigma_0(i)^{mu(i)}prod_{i=1}^{n}sigma_0(i)^{i}
]
考虑 (prod_{i=1}^{n}sigma_0(i)^{mu(i)})
运用 (mu) 的性质，设 (c(i)) 表示 (i) 的质数因子个数
那么就是
[prod_{i=1}^{n}2^{mu(i)c(i)}=2^{sum_{i=1}^{n}mu(i)c(i)}
]
只需要设
(f(x,j)) 表示 (1) 到 (x) 最小质因子大于等于第 (j) 个质数的合数的 (mu(i)c(i)) 的和
(g(x,j)) 表示 (1) 到 (x) 最小质因子大于等于第 (j) 个质数的合数的 (mu(i)) 的和
预处理出 (cntp(i)) 表示 (1) 到 (x) 的质数个数就可以递推了
这一部分直接 (min25) 筛即可
考虑 (prod_{i=1}^{n}sigma_0(i)^{i})
分开考虑每个质数的贡献
设 (s(n,p,k)) 表示 (1) 到 (n) 中只含有 (p^k) 这个因子不含 (p^i,i
e k) 的数的和
设 (P) 为质数集合
那么
[prod_{i=1}^{n}sigma_0(i)^{i}=prod_{pin P}prod_{k=1}(k+1)^{s(n,p,k)}
]
设 (Sum(x)=sum_{i=1}^{x}i)
那么
[s(n,p,k)=p^kSum(lfloorfrac{n}{p^k}
floor)-p^{k+1}Sum(lfloorfrac{n}{p^{k+1}}
floor)
]
当 (ple sqrt{n}) 的时候直接暴力计算
当 (p > sqrt{n}) 的时候，显然 (k) 最多就是 (1)
那么
[s(n,p,k)=pSum(lfloorfrac{n}{p}
floor)
]
(min25) 筛预处理出 (S(x)) 表示 (1) 到 (x) 的质数的和，然后数论分块即可
注意常数优化QwQ
# include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod(1e12 + 39);
const ll phi(1e12 + 38);
const int maxn(1e6 + 5);
inline ll Mul1(ll x, ll y) {
register ll ret = x * y - (ll)((long double)x / mod * y + 0.5) * mod;
return ret < 0 ? ret + mod : ret;
}
inline ll Mul2(ll x, ll y) {
register ll ret = x * y - (ll)((long double)x / phi * y + 0.5) * phi;
return ret < 0 ? ret + phi : ret;
}
inline ll Pow(ll x, ll y) {
register ll ret = 1;
for (; y; y >>= 1, x = Mul1(x, x))
if (y & 1) ret = Mul1(ret, x);
return ret;
}
inline void Inc1(ll &x, ll y) {
x = x + y >= mod ? x + y - mod : x + y;
}
inline void Inc2(ll &x, ll y) {
x = x + y >= phi ? x + y - phi : x + y;
}
inline ll Dec1(ll x, ll y) {
return x - y < 0 ? x - y + mod : x - y;
}
inline ll Dec2(ll x, ll y) {
return x - y < 0 ? x - y + phi : x - y;
}
int test, pr[maxn], tot, d, id1[maxn], id2[maxn], cnt;
ll n, val[maxn], cntp[maxn], f[maxn], g[maxn], s[maxn], sp[maxn];
bitset <maxn> ispr;
# define ID(x) ((x) <= d ? id1[x] : id2[n / (x)])
inline ll Sum1(ll x) {
return (x & 1) ? Mul1(x, (x + 1) / 2) : Mul1(x / 2, x + 1);
}
inline ll Sum2(ll x) {
return (x & 1) ? Mul2(x, (x + 1) / 2) : Mul2(x / 2, x + 1);
}
inline ll GetSum(ll x, ll v1, ll v2) {
return Dec2(Mul2(v1, Sum2(x / v1)), Mul2(v2, Sum2(x / v2)));
}
inline void Solve() {
while (cnt) f[cnt] = g[cnt] = 0, cnt--;
register ll i, j, id, ans1, ans2, v, ret, lst, cur;
for (d = sqrt(n), i = 1; i <= n; i = j + 1) {
j = n / (n / i), val[++cnt] = n / i;
val[cnt] <= d ? id1[val[cnt]] = cnt : id2[j] = cnt;
cntp[cnt] = val[cnt] - 1, s[cnt] = Sum2(val[cnt]) - 1;
}
for (i = 1; i <= tot && pr[i] <= n / pr[i]; ++i)
for (j = 1; j <= cnt && pr[i] <= val[j] / pr[i]; ++j) {
id = ID(val[j] / pr[i]), cntp[j] -= cntp[id] - i + 1;
Inc2(s[j], Mul2(pr[i], Dec2(sp[i - 1], s[id])));
}
for (--i; i; --i)
for (j = 1; j <= cnt && pr[i] <= val[j] / pr[i]; ++j) {
id = ID(val[j] / pr[i]);
v = Dec2(cntp[id] - i, g[id]), Inc2(g[j], v);
Inc2(v, cntp[id] - i), Inc2(f[j], Dec2(v, f[id]));
}
for (i = 1; i <= cnt; ++i) Inc2(f[i], phi - cntp[i]);
ans1 = Pow(2, f[ID(n)]), ans2 = 1;
for (i = 1; i <= tot && pr[i] <= n / pr[i]; ++i)
for (j = 1, v = pr[i]; v <= n; v = v * pr[i], ++j) ans2 = Mul1(ans2, Pow(j + 1, GetSum(n, v, v * pr[i])));
for (lst = sp[i - 1], ret = 0, i = pr[i]; i <= n; i = j + 1) {
j = n / (n / i);
Inc2(ret, Mul2(Dec2(cur = s[ID(j)], lst), Sum2(n / i))), lst = cur;
}
ans2 = Mul1(ans2, Pow(2, ret)), printf("%lld
", Mul1(ans1, ans2));
}
int main() {
register int i, j;
ispr[1] = 1;
for (i = 1; i < maxn; ++i) {
if (!ispr[i]) pr[++tot] = i, sp[tot] = (sp[tot - 1] + i) % phi;
for (j = 1; j <= tot && i * pr[j] < maxn; ++j) {
ispr[i * pr[j]] = 1;
if (!(i % pr[j])) break;
}
}
scanf("%d", &test);
while (test) scanf("%lld", &n), Solve(), --test;
return 0;
}