HDU 6322(打表)
Problem D. Euler Function
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 138 Accepted Submission(s): 124
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
打了個表,發現除了1 2 3 4 6 不符合條件,其他都是
#include <bits/stdc++.h> #define fir first #define se second #define mp make_pair #define pb push_back #define ll long long using namespace std; const int maxn=3000+10; const ll mod=998244353; const double eps=1e-7; const int maxm=1e6+10; const int inf=0x3f3f3f3f; int phi(int x){ int res=x; for (int i=2;i*i<=x;i++){ if (x%i==0){ while (x%i==0) x/=i; res/=i; res*=(i-1); } } if (x>1){ res/=x; res*=(x-1); } return res; } int check(int x){ int res=phi(x); for (int i=2;i<res;i++){ if(res%i==0) return 1; } return 0; } int ans[5]={5,7,8,9,10}; int main(){ int t; scanf("%d",&t); while (t--){ int k; scanf("%d",&k); if (k<=5){ printf("%d\n",ans[k-1]); continue; } else{ int dif=(k-5)+10; printf("%d\n",dif); } } return 0; }