SPOJ AMR11E Distinct Primes 基础数论

Arithmancy is Draco Malfoy's favorite subject, but what spoils it for him is that Hermione Granger is in his class, and she is better than him at it. Prime numbers are of mystical importance in Arithmancy, and Lucky Numbers even more so. Lucky Numbers are those positive integers that have at least three distinct prime factors; 30 and 42 are the first two. Malfoy's teacher has given them a positive integer n, and has asked them to find the n-th lucky number. Malfoy would like to beat Hermione at this exercise, so although he is an evil git, please help him, just this once. After all, the know-it-all Hermione does need a lesson.

Input

The first line contains the number of test cases T. Each of the next T lines contains one integer n.

Output

Output T lines, containing the corresponding lucky number for that test case.

Constraints

1 <= T <= 20
1 <= n <= 1000

Example

Sample Input:
2
1
2

Sample Output:
30
42

题意：找第n个由至少三个不同素因子组成的数。

思路：n<=1000直接暴力打表预处理

/** @Date    : 2016-12-11-19.01
  * @Author  : Lweleth (SoungEarlf@gmail.com)
  * @Link    : https://github.com/
  * @Version :
  */
#include<bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

int pri[N];
int ans[10010];
int c = 0;
bool vis[N];
int prime()
{
    MMF(vis);
    for(int i = 2; i < N; i++)
    {
        if(!vis[i])
            pri[c++] = i;

        for(int j = 0; j < c && pri[j]*i < N; j++)
        {
            vis[i * pri[j]] = 1;
            if(i % pri[j] == 0)
                break;
        }
    }
}

int main()
{
    prime();
    int tot = 0;
    for(int i = 0; i <= 10000; i++)
    {
        int t = i;
        int cnt = 0;
        for(int j = 0; j < c && pri[j]*pri[j] <= t; j++)
        {
            if(t % pri[j] == 0)
            {
                cnt++;
                while(t % pri[j] == 0)
                    t /= pri[j];
            }
        }
        if(t > 1)
            cnt++;
        if(cnt >= 3)
            ans[tot++] = i;
    }
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n;
        scanf("%d", &n);
        printf("%d\n", ans[n - 1]);
    }
    return 0;
}