Description

Time Limit: 5000 MS    Memory Limit: 65536 K 
Description
A positive integer is said to be squarefree if it is divisible by no perfect square larger than 1. 
For example, the first few squarefree numbers are {1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, ...}. 
Can you tell me the K-th ( 1-indexed ) smallest squarefree number. 

Input
The first line of the input will be a integer to represent the number of test cases.
For each test case there is only one line contains only one integer K. 
( 1 <= K <= 10^9 )
There is a blank line before each test case.

Output
For each test case output the answer on a single line:
The K-th smallest squarefree number.

Sample Input
6

1

2

7

1000

1234567

1000000000 
Sample Output
1
2
10
1637
2030745
1644934081 
Source

#include<iostream>
#include<algorithm>
using namespace std;

typedef long long ll;
const int MAXN = 4000;
const int MAXM = 100100;

int isprime[MAXN];
int prime[MAXN];
int co[MAXM];
int cnt;

void getP()  //线性素数筛
{
	for(int i = 1; i < MAXN; i++)
		isprime[i] = 1;
	for(int i = 2; i < MAXN; i++)
	{
		if(isprime[i])
			prime[cnt++] = i;
		for(int j = 0; j < cnt&&i*prime[j] < MAXN; j++)
		{
			isprime[i*prime[j]] = 0;
			if(i%prime[j] == 0)
				break;
		}
	}
}

void getCo() //计算素数矩阵
{
	for(int i = 2; i < MAXM; i++)
	{
		co[i] = 1; 
		int tem = i;
		for(int j = 0; prime[j] * prime[j] <= tem; j++)
		{
			if(tem%prime[j] == 0)
			{
				co[i] = -co[i];		//找到一个质因子则改变容斥时的加减性
				tem /= prime[j];
			}
			if(tem%prime[j] == 0)
			{
				co[i] = 0;		//不在我们的考虑范围内
				break;
			}
		}
		if(tem > 0&&co[i]!=0) co[i] = -co[i];
	}
}

int main()
{
	int kase, n;
	cin >> kase;
	getP();
	getCo();
	while(kase--)
	{
		cin >> n;
		ll l = 1, r = 2e10;
		ll mid, tem,ans;
		while(l <= r)  //二分数，找到第k个数
		{
			mid = l + (r - l) / 2;
			tem = mid;
			for(long long i = 2; i*i <= mid; i++)
			{
				tem = tem + (mid / (i*i))*co[i];  //mid/(i*i)算出在mid范围内有多少个数能被(i*i)整除，再乘上容斥系数
			}

			if(tem >= n)
			{
				r = mid - 1;
				ans = mid;
			}
			else	l = mid + 1;
		}
		cout << ans << endl;

	}
	return 0;
}