hdu6322（欧拉函数）

Problem D. Euler Function

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 46    Accepted Submission(s): 46

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 1itself, 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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题意：求第k小的数，满足φ(n)是合数。φ(n)是欧拉函数。

解析：当n>=8时，φ(n)恒定为合数。其他的特判就可以。

#include
using namespace std;
 
#define e exp(1)
#define pi acos(-1)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
#define mem(a,b) memset(a,b,sizeof(a))
int gcd(int a,int b){return b?gcd(b,a%b):a;}

int main()
{
    int T;scanf("%d",&T);
    while(T--)
    {
        int k;scanf("%d",&k);
        if(k==1)puts("5");
        else if(k==2)puts("7");
        else printf("%d\n",k+5);
    }
    return 0;
}