NYOJ 939最小公倍数

/**
*    求LCM(1,2,3,...n)
**/

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
#define N 100000010

unsigned int prime[6000000], p[6000000], T, num;
unsigned int flag[3126000];

void slove(int n) {
    unsigned int res = 1;
    int i, t1, t2;
    for(i = 0; i <= num && prime[i] * prime[i] <= n; ++i) {
        t1 = prime[i];
        t2 = prime[i] * prime[i];
        while(t2 / t1 == prime[i] && t2 <= n)
        {
            t1 *= prime[i];
            t2 *= prime[i];
        }
        res *= (t1 / prime[i]);
    }
    int t = upper_bound(prime, prime+num, n) - prime - 1;
    res *= p[t];
    printf("%u
", res);
}

//生成素数
void initPrime() {
    int i, j;
    p[num=0] = prime[0] = 2;
    for(i = 3; i < N; i += 2) {
        if( !(flag[i/32]&(1<<(i%32))) ) {
            prime[++num] = i;
            p[num] = i * p[num-1];
            for(j = i * 3; j < N; j += i * 2)
                flag[j/32] |= 1<<(j%32);
        }
    }
}

int main() {
    initPrime();
    prime[++num] = 100000003;
    scanf("%d", &T);; int i, n;
    for(i = 0; i < T; ++i) {
        scanf("%d", &n);
        slove(n);
    }
    return 0;
}