HDU 6322 Euler Function

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 题目大意：介绍了一下欧拉函数的定义与合数的定义（欧拉函数：https://baike.baidu.com/item/欧拉函数/1944850?fr=aladdin  合数：https://baike.baidu.com/item/合数/49186）然后要求输出欧拉函数中的第k个合数。 思路：这道题是签到题，看完题第一反应是打表直接找，但是，这样肯定是超时的(我做题一般都不怎么带脑子)，所以应该找规律，从合数的定义可以得出，，，，只要是能被2整除且不等于2的欧拉数都是合数，，然后回头瞟一眼欧拉函数前几十项，，，除了前几项全是偶数且不等于2.............. 代码如下： #include<stdio.h> int main() { int t,a; scanf("%d",&t); while(t--) { scanf("%d",&a); if(a==1) printf("5\n"); else printf("%d\n",a+5); } return 0; } ​​