2018HDU多校训练-3-Problem D. Euler Function
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链接:http://acm.hdu.edu.cn/showproblem.php?pid=6322
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.
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) .
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
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题解:给你一个数k,让你求让你求第k 个gcd(num,x)的个数为合数(除了1)的num,x为从1 ~ num-1,这题题名写着欧拉函数,很明显让你求第k个欧拉函数值为合数的数;
显然,由于大于3的质数都满足题意(根据欧拉函数知道,质数的欧拉函数值为x-1,必为大于2的偶数)
对于奇数: 有当m,n互质时,有f(mn)=f(m)f(n),根据任何数都可以由多个质数的多少次幂相乘得到,故,对于质数num,其可以由一个质数乘另一个数得到,质数和任意数都是互质的,故f(num)=f(x)f(y){假设x为质数},则,f(num)=(x-1)*f(y),由(x-1)为偶数,且f(y)>1,则对于任意奇数都是满足题意的;
对于偶数:由上同理可以推出只有6不满足题意:故只要排除6即可;从4开始遍历:
参考代码为:
#include<bits/stdc++.h> using namespace std; int main() { int t; long long k; cin>>t; while(t--) { cin>>k; if(k==1) cout<<5<<endl; else cout<<k+5<<endl; } return 0; }
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