A - Bi-shoe and Phi-shoe (欧拉函数打表)

Description

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

3

5

1 2 3 4 5

6

10 11 12 13 14 15

2

1 1

Sample Output

Case 1: 22 Xukha

Case 2: 88 Xukha

Case 3: 4 Xukha

题意 ： 给你 n 个数，分别找出大于等于这些数的欧拉函数值的 x ,使 x 的和最小

//我的思路：首先欧拉函数打表，再分别找出最小值

//之前纠结于 1 1 得出 4 的这组数据 后来看见 Φ (n) = numbers less than n which are relatively prime

//这个题中的规定和欧拉函数不一样 欧拉函数是小于等于 n , 所以在本题中 1 的值 不为 1 而是 0；

#include <iostream>
#include <algorithm>
#include <cstdio>
#define ll long long
using namespace std;
#define max 1123456

ll eular[max];

void init()
{
    eular[]=;
    for(int i=;i<max;i++)
     eular[i]=i;
    for(int i=;i<max;i++)
    {
        if(eular[i]==i)
        {
            for(int j=i;j<max;j+=i)
            eular[j]=eular[j]/i*(i-);
        }
    }
}

int main()
{
    ll c,data[],ans,tmp=;
    cin>>c;
    init();
    while(c--)
    {
        int n;
        tmp++;
        cin>>n;
        ans=;
        for(int i=;i<n;i++)
        cin>>data[i];
        for(int i=;i<n;i++)
        {
            for(int j=data[i]+;j<max;j++)
            if(eular[j]>=data[i])
            {
                ans+=j;
                break;
            }
        }
        printf("Case %lld: %lld Xukha\n",tmp,ans);
    }
    return ;
}