2018-07-31 | Contest 3 HDU 6322 Euler Function

Problem D. Euler Function

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 198    Accepted Submission(s): 175

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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题目意思是：返回第k个欧拉函数φ(n)结果为复合数的数（与欧拉所得值大小无关）

打表找规律，可发现从8开始后面φ(n)结果全为大于2的偶数，都为复合数

φ(n)	0	1	2	3	4	5	6	7	8	9
0?	0	1	1	2	2	4	2	6	4	6
1?	4	10	4	12	6	8	8	16	6	18
2?	8	12	10	22	8	20	12	18	12	28
3?	8	30	16	20	16	24	12	36	18	24
4?	16	40	12	42	20	24	22	46	16	42
5?	20	32	24	52	18	40	24	36	28	58
6?	16	60	30	36	32	48	20	66	32	44
7?	24	70	24	72	36	40	36	60	24	78
8?	32	54	40	82	24	64	42	56	40	88
9?	24	72	44	60	46	72	32	96	42	60

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欧拉函数代码：（来源）：https://blog.youkuaiyun.com/once_hnu/article/details/6302868

//直接求解欧拉函数  
int euler(int n){ //返回euler(n)   
     int res=n,a=n;  
     for(int i=2;i*i<=a;i++){  
         if(a%i==0){  
             res=res/i*(i-1);//先进行除法是为了防止中间数据的溢出   
             while(a%i==0) a/=i;  
         }  
     }  
     if(a>1) res=res/a*(a-1);  
     return res;  
}  
  
//筛选法打欧拉函数表   
#define Max 1000001  
int euler[Max];  
void Init(){   
     euler[1]=1;  
     for(int i=2;i<Max;i++)  
       euler[i]=i;  
     for(int i=2;i<Max;i++)  
        if(euler[i]==i)  
           for(int j=i;j<Max;j+=i)  
              euler[j]=euler[j]/i*(i-1);//先进行除法是为了防止中间数据的溢出   
}  

ac代码：

#include<bits/stdc++.h>
using namespace std;
int main(){
    int T,n;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        if(n == 1){
            puts("5");
        }else if(n == 2){
            puts("7");
        }else{
            printf("%d\n",n + 5);
        }
    }
    return 0;
}