https://vjudge.net/contest/324284#problem/B 数学水题，其实就是想写下位图。。和状压很像 题意：给n让求lcm(1,2,3,...,n)，n=1e8 思路：显然ans = 所有小于n的素数p[i]的max(p[i]^k)相乘。由于空间太大，
https://vjudge.net/contest/324284#problem/B
数学水题，其实就是想写下位图。。和状压很像
题意：给n让求lcm(1,2,3,...,n)，n<=1e8
思路：显然ans = 所有小于n的素数p[i]的max(p[i]^k)相乘。由于空间太大，装素数的数组开不下，要用位图，int可以保存32位二进制，我们可以把每一位当作一个数，又因为偶数除了2以外都不是素数，所以只需筛选奇数。
L(1) = 1
L(x+1) = { L(x) * p if x+1 is a perfect power of prime p
{ L(x) otherwise
L(2) = 1 * 2
L(3) = 1 * 2 * 3
L(4) = 1 * 2 * 3 * 2 // because 4 = 2^2
L(5) = 1 * 2 * 3 * 2 * 5
L(6) = 1 * 2 * 3 * 2 * 5 // 6 is not a perfect power of a prime
L(7) = 1 * 2 * 3 * 2 * 5 * 7
#include <bits/stdc++.h>
using namespace std;
typedef unsigned int UI;
const int maxn = 100000005;
const int N = 5800000;
UI mul[N];
int vis[maxn/32+10], p[N];
int cnt, n;
void init ()
{
cnt = 1;
p[0] = mul[0] = 2;
for (int i=3; i<maxn; i+=2)
if (!(vis[i/32]&(1<<((i/2)%16))))
{//寻找代表i的哪一位，偶数不占位数
p[cnt] = i;
mul[cnt] = mul[cnt-1] * i;
for (int j=3*i; j<maxn; j+=2*i)
vis[j/32] |= (1<<((j/2)%16));//删除有因子的位数
cnt ++;
}
//printf ("%d\n", cnt);
}
UI solve ()
{
int pos = upper_bound(p, p+cnt, n) - p - 1;//找出最大的比n小的素数
UI ans = mul[pos];
for (int i=0; i<cnt&&p[i]*p[i]<=n; i++)
{
int tem = p[i];
int tt = p[i] * p[i]; //这个tt很有可能溢出int
while (tt/tem == p[i]&&tt<=n)
{
tem *= p[i];
tt *= p[i];
}
ans *= tem / p[i];
}
return ans;
}
int main ()
{
int t, l = 0;
init ();
scanf ("%d", &t);
while (t --)
{
scanf ("%d", &n);
printf ("Case %d: %u\n", ++l, solve());
}
return 0;
}