https://vjudge.net/contest/324284#problem/B
Water math problem, in fact, would like to write a bitmap. . And shaped like pressure
Meaning of the questions: to let n seek lcm (1,2,3, ..., n), n <= 1e8
Thinking: Obviously ans = n less than all primes p [i] is max (p [i] ^ k) is multiplied. Because space is too large, the number of array elements installed no less than open, use bitmaps, int can save 32-bit binary, we can each as a number, but also because in addition to the even number 2 is not a prime number, so just odd screening .
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.>
the using namespace STD;
typedef unsigned int the UI;
const int MAXN = 100 000 005;
const int N = 5.8 million;
the UI MUL [N];
int VIS [MAXN / 32 + 10], P [N] ;
int CNT, n-;
void the init ()
{
CNT =. 1;
P [0] = MUL [0] = 2;
for (int I =. 3; I <MAXN; I + = 2)
! IF ((VIS [I / 32] & (<<. 1 ((i / 2)% 16))))
{// find where i is a representative, not accounting for the even bits
P [CNT] = i;
MUL [CNT] = MUL [CNT -1] * I;
for (int. 3 J = I *; J <MAXN; = 2 * I + J)
VIS [J / 32] | = (. 1 << ((J / 2)% 16)); // delete the factor of the number of bits
cnt ++;
}
// printf ( "% d \ the n-", cnt);
}
UI solve ()
{
Int POS = upper_bound, (P, P + CNT, n) - P -. 1; // find the maximum is smaller than n prime
the 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]; // this is likely to overflow tt int
the while (tt / TEM == P [I] && TT <= n-)
{
TEM * = P [I];
TT * = P [I];
}
ANS * = TEM / P [I];
}
return ANS;
}
int main ()
{
int T, 0 = L;
the init ();
Scanf ( "% D", & T);
the while (T -)
{
Scanf ( "% D", & n-);
the printf ( "% Case D: U% \ n-", L ++, Solve ());
}
return 0;
}