【欧拉函数】【二分】【欧拉函数模板】

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在竹竿竞速运动中，Phi教练需要为学生购买竹竿。每根竹竿的得分基于其长度与费马数的关系。Bi助手需在预算内为每位学生选择得分不低于其幸运数字的竹竿。本文探讨了如何高效地解决此问题。

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1370 - Bi-shoe and Phi-shoe

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Time Limit: 2 second(s)

Memory Limit: 32 MB

Bamboo Pole-vault is a massively popular sport in Xzhiland. And Master Phi-shoe is a very popular coach for his success. He needs some bamboos for his students, so he asked his assistant Bi-Shoe to go to the market and buy them. Plenty of Bamboos of all possible integer lengths (yes!) are available in the market. According to Xzhila tradition,

Score of a bamboo = Φ (bamboo's length)

(Xzhilans are really fond of number theory). For your information, Φ (n) = numbers less than n which are relatively prime (having no common divisor other than 1) to n. So, score of a bamboo of length 9 is 6 as 1, 2, 4, 5, 7, 8 are relatively prime to 9.

The assistant Bi-shoe has to buy one bamboo for each student. As a twist, each pole-vault student of Phi-shoe has a lucky number. Bi-shoe wants to buy bamboos such that each of them gets a bamboo with a score greater than or equal to his/her lucky number. Bi-shoe wants to minimize the total amount of money spent for buying the bamboos. One unit of bamboo costs 1 Xukha. Help him.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 10000) denoting the number of students of Phi-shoe. The next line contains n space separated integers denoting the lucky numbers for the students. Each lucky number will lie in the range [1, 10⁶].

Output

For each case, print the case number and the minimum possible money spent for buying the bamboos. See the samples for details.

Sample Input

Output for Sample Input

1 2 3 4 5

10 11 12 13 14 15

1 1

Case 1: 22 Xukha

Case 2: 88 Xukha

Case 3: 4 Xukha

#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;
    
int euler[2000001];
void EU()
{
	euler[1] = 1;
	for(int i=2;i<=2000000;i++)
	{
		if(!euler[i])
		{
			for(int j=i;j<=2000000;j+=i)
			{
				if(!euler[j])
					euler[j] = j;
				euler[j] = euler[j] / i * (i - 1);
			}
		}
	}
}
int main()
{
  
    
	EU();
	for(int i=1;i<=2000000;i++) euler[i] = max(euler[i-1],euler[i]);
	euler[1] = 0;
	int T;
	scanf("%d",&T);
	int C = 1;
	while(T--)
	{
		int n,x;
		long long ans = 0;
		scanf("%d",&n);
		while(n --)
		{
			scanf("%d",&x);
			ans += lower_bound(euler,euler+2000000,x) - euler;
		}
		printf("Case %d: %I64d Xukha\n",C++,ans);
	}
 
    return 0;
}