Problem D. Euler Function

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Problem D. Euler Function

题目:

 

Problem D. Euler Function

http://acm.hdu.edu.cn/showproblem.php?pid=6324

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 66    Accepted Submission(s): 64


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 1kn 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 nis 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(1T100000), denoting the number of test cases.
In each test case, there is only one integer k(1k109).
 

 

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

 

 

题意:

  给定k, 求第k小的数n, 满足 φ(n)是合数

 

思路

   暴力打表后发现除了,φ(i) i=1..5,6 不是合数外,其余都是,因而 f(1)=5,  f(k)=k+5 (k>=2)

 

证明

技术分享图片

技术分享图片

 

技术分享图片

 

 技术分享图片

 

代码:

#include<stdio.h>int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        if(n==1)printf("5
");
        else printf("%d
",n+5);
    }
    return 0;
}

 












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