[洛谷P4626]一道水题 II

题目大意：求$lcm(1,2,3,\cdots,n)\pmod{100000007}$，$n\leqslant10^8$

题解：先线性筛出质数，然后求每个质数最多出现的次数，可以用$\log_in$来求，$i$为该质数。使用换底公式$log_in=\dfrac{\log_2n}{\log_2i}$。

卡点：模数是$10^8+7$，看成$10^9+7$

C++ Code：

#include <algorithm>
#include <bitset>
#include <cstdio>
#include <cmath>
const int mod = 1e8 + 7;
inline int pw(int base, int p) {
static int res;
for (res = 1; p; p >>= 1, base = static_cast<long long> (base) * base % mod) if (p & 1) res = static_cast<long long> (res) * base % mod;
return res;
}std::bitset<100000010> npri;
int plist[5761460], ptot;
void sieve(const int n) {
for (register int i = 2; i <= n; ++i) {
if (!npri[i]) plist[ptot++] = i;
for (register int j = 0, t; (t = plist[j] * i) <= n; ++j) {
npri.set(t);
if (i % plist[j] == 0) break;
}
}
}int n;
long long ans = 1;
int main() {
scanf("%d", &n);
sieve(n);
for (register int i = 0; i < ptot; ++i) {
ans = ans * pw(plist[i], static_cast<int> (log2(n) / log2(plist[i]))) % mod;
}
printf("%lld\n", ans);
return 0;
}