LightOJ - 1370 Bi-shoe and Phi-shoe （欧拉函数）

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 nwhich 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、

PS：典型的欧拉函数，因为后面要加，所以要提前预处理一下。

#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<stack>
#include<string>
const int maxn=1e6+10;
const int mod=1e9+7;
const int inf=1e9;
#define me(a,b) memset(a,b,sizeof(a))
typedef long long ll;
using namespace std;
int a[maxn];
void inct()
{
    me(a,0);
    for(int i=2;i<maxn;i++)
        if(!a[i])
        {
            for(int j=i;j<maxn;j+=i)
            {
                if(!a[j])
                    a[j]=j;
                a[j]=a[j]/i*(i-1);
            }
        }
}
int main()
{
    int t,Case=1;cin>>t;
    inct();
    while(t--)
    {
        int n;cin>>n;
        ll sum=0;
        for(int i=0;i<n;i++)
        {
            int x;scanf("%d",&x);
            for(int j=x+1;j<maxn;j++)
                if(a[j]>=x)
                {
                    sum+=j;
                    break;
                }
        }
        cout<<"Case "<<Case++<<": "<<sum<<" Xukha"<<endl;
    }
    return 0;
}