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 Input
Case 1: 22 Xukha
Case 2: 88 Xukha
Case 3: 4 Xukha

题意：每一个人都有一个幸运数，那么他的鞋的价值就是大于等于他的幸运数的欧拉数的数。求n个人的最小的花费。直接打表。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int Max = 1e6+10;

int phi[Max];

void Init()
{
    memset(phi,0,sizeof(phi));

    phi[1] = 1;

    for(int  i = 2;i < Max; i++)
    {
        if(!phi[i])
        {
            for(int j = i;j < Max ; j+=i)
            {
                if(!phi[j]) phi[j] = j;

                phi[j] = phi[j]/i*(i-1);
            }
        }

    }
}

int a[Max];

int main()
{
    int T,z = 1;

    int n;

    Init();

    scanf("%d",&T);

    while(T--)
    {
        scanf("%d",&n);

        for(int i = 0;i < n; i++) scanf("%d",&a[i]);

        sort(a,a+n);

        LL ans = 0 ;

        for(int i = 0; i < n ; i++)
        {
            int j = a[i]+1;

            while(phi[j]<a[i]) j++;

            ans+=j;
        }

        printf("Case %d: %lld Xukha\n",z++,ans);
    }
    return 0;
}