HDU-6322 Problem D. Euler Function

Problem D. Euler Function

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 121    Accepted Submission(s): 111

 

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

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define clr(a) memset(a,0,sizeof(a))

const int MAXN = 1e5+10;
const int INF = 0x3f3f3f3f;
const int N = 1010;

int t,n;
int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        int ans = 0;
        if(n==1)    ans = 5;
        else ans = (n-2)+7;
        cout<<ans<<endl;
    }
    return 0;
}