标签：
Description
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 nth 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 (STDIN):
The first line contains the number of test cases T. Each of the next T lines contains one integer n.
Output (STDOUT):
Output T lines, containing the corresponding lucky number for that test case.
Constraints:
1 <= T <= 20
1 <= n <= 1000
Sample Input:
2
1
2
Sample Output:
30
42
Hint
| Added by: | Varun Jalan |
| Date: | 2011-12-15 |
| Time limit: | 3s |
| Source limit: | 50000B |
| Memory limit: | 1536MB |
| Cluster: | Cube (Intel G860) |
| Languages: | All |
| Resource: | Varun Jalan - ICPC Asia regionals, Amritapuri 2011 |
Source
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=121703#problem/E
My Solution
分解质因数， 至少有三个不同的质因数的数是lucky number, 用修改分解质因数的模版， 是只记录质因数个数就好，不用管是什么而且不同质因数个数大于等于3就return 3
用的分解质因素模版不知道为什么会WA， 改成 2~n, 暴力试除来分解质因数倒是可以⊙﹏⊙‖∣
原因， 对于2~sqrt(n)+1， 可能有剩余的大于 sqrt(n) +1的素数， 也就是还有大于sqrt(n)+1 的质因数， 所以试除全部的2~n比较好
//刚开始做的时候， 看成了，只能有三个不同质数构成的数， 然后就搞出个素数表，然后构造，然后用小根堆保存， 然后从那10000000多个根据要求的数中拿出最小的1000多个放到thisans[]里
//(┬＿┬)成功浪费了一大堆时间
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long LL;
const int maxn = 20000 + 9;
int ans[maxn];
int factor(int n)
{
int tot = 0;
int temp, i, now;
//temp = (int)((double)sqrt(n) + 1);
now = n;
for(int i = 2; i <= n; i++){ //!!!!!!为什么不能用temp呢
if(now%i == 0){
tot++;
if(tot >= 3) return tot;
}
while(now%i == 0){
now/=i;
}
}
return tot;
}
int main()
{
int t = 1, num = 2;
while(t < 1009){
if(factor(num) >= 3) {ans[t] = num; t++;}
num++;
}
//cout<<factor(4588)<<endl;
#ifdef LOCAL
freopen("a.txt", "r", stdin);
//freopen("b.txt", "w", stdout);
#endif // LOCAL
LL T, n;
cin>>T;
while(T--){
cin>>n;
cout<<ans[n]<<endl;
}
return 0;
}
Thank you!
------from ProLightsUESTC 2016 Summer Training #2 Div.2 E 分解质因素(除了以后剩下的可能也是个素数)
标签：
原文地址：http://blog.csdn.net/prolightsfxjh/article/details/51893130
