链接：http://acm.hdu.edu.cn/showproblem.php?pid=6322
Problem Description
In number theory, Euler‘s totient function φ(n) counts the positive integers up to a given integer n
that are relatively prime to n
. It can be defined more formally as the number of integers k
in the range 1≤k≤n
for which the greatest common divisor gcd(n,k)
is equal to 1
.
For example, φ(9)=6
because 1,2,4,5,7
and 8
are coprime with 9
. As another example, φ(1)=1
since for n=1
the only integer in the range from 1
to n
is 1
itself, and gcd(1,1)=1
.
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 1
and itself. So obviously 1
and all prime numbers are not composite number.
In this problem, given integer k
, your task is to find the k
-th smallest positive integer n
, that φ(n)
is a composite number.
Input
The first line of the input contains an integer T(1≤T≤100000)
, denoting the number of test cases.
In each test case, there is only one integer k(1≤k≤109
)
.
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;
}
原文地址：https://www.cnblogs.com/songorz/p/9398398.html