2018HDU多校训练-3-Problem D. Euler Function

链接：http://acm.hdu.edu.cn/showproblem.php?pid=6322

Problem Description

In number theory, Euler's totient function

because

and

are coprime with

. As another example,

since for

the only integer in the range from

itself, and

.
A composite number is a positive integer that can be formed by multiplying together two smaller positive integers. Equivalently, it is a positive integer that has at least one divisor other than

and itself. So obviously

and all prime numbers are not composite number.
In this problem, given integer

, your task is to find the

-th smallest positive integer

, that

is a composite number.

Input

The first line of the input contains an integer

, denoting the number of test cases.
In each test case, there is only one integer

Output

For each test case, print a single line containing an integer, denoting the answer.

Sample Input

2 1 2

Sample Output

5 7

Source

2018 Multi-University Training Contest 3

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题解：给你一个数k，让你求让你求第k 个gcd(num,x)的个数为合数（除了1）的num,x为从1 ~ num-1,这题题名写着欧拉函数，很明显让你求第k个欧拉函数值为合数的数；

显然，由于大于3的质数都满足题意（根据欧拉函数知道，质数的欧拉函数值为x-1,必为大于2的偶数）

对于奇数：有当m,n互质时，有f(mn)=f(m)f(n)，根据任何数都可以由多个质数的多少次幂相乘得到，故，对于质数num,其可以由一个质数乘另一个数得到，质数和任意数都是互质的，故f(num)=f(x)f(y){假设x为质数}，则，f(num)=(x-1)*f(y),由(x-1)为偶数，且f(y)>1,则对于任意奇数都是满足题意的；

对于偶数：由上同理可以推出只有6不满足题意：故只要排除6即可；从4开始遍历：

参考代码为：

#include<bits/stdc++.h>
using namespace std;

int main()
{
	int t;
	long long k;
	cin>>t;
	while(t--)
	{
		cin>>k;
		if(k==1) cout<<5<<endl;
		else cout<<k+5<<endl;
	}
	return 0;
}