P4626 一道水题 II
求 $lcm[1, \ldots n]$ 。筛出质数，考虑每个质数的最大指数。
注意到 $n$ 很大，空间限制不能直接开 $10 ^ 8$ 个 int ，所以用 bitset 。
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| #include <cstdio>
#include <bitset>
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
if (flag) x = ~x + 1;
}
template <class Type, class ...rest>
void read(Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write(Type x)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
typedef long long LL;
const int N = 1e8 + 10, M = 5761456 + 10, mod = 1e8 + 7;
bitset <N> vis;
int n, cnt, p[M];
void init ()
{
vis[1] = true;
for (int i = 2; i <= n; ++i)
{
if (!vis[i]) p[++cnt] = i;
for (int j = 1; j <= cnt && (LL)i * p[j] <= n; ++j)
{
vis[i * p[j]] = true;
if (i % p[j] == 0) break;
}
}
}
int main ()
{
read(n);
init();
int res = 1;
for (int i = 1; i <= cnt; ++i)
{
int t = p[i];
while ((LL)t * p[i] <= n) t *= p[i];
res = (LL)res * t % mod;
}
write(res), puts("");
return 0;
}
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