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：Φ (n)表示长度为小于数字n的和n互质的数的个数，也就是欧拉函数。现在给出n个幸运数字，对于每一个幸运数字，要求的x，使Φ (x)的值大于等于这个幸运数字，求这些x和的最小值
知识点：素数打表（埃氏筛法），欧拉函数的性质
暂时还没出证明会不会存在x不是素数也能使题意成立，所以这样的解法还不能让我信服
#include<iostream>
#define maxn 1000010
using namespace std;
int T;
int n,num;
int p[maxn]={0};
void find_prime(){
for(int i=2;i<maxn;i++){
if(p[i]==false){
for(int j=i+i;j<maxn;j=j+i){
p[j]=true;
}
}
}
}
int main(){
cin>>T;
find_prime();
for(int i=1;i<=T;i++){
cin>>n;
long long int ans=0;
for(int j=0;j<n;j++){
cin>>num;
for(int k=num+1;k<maxn;k++){
if(p[k]==false){
ans+=k;
break;
}
}
}
cout<<"Case "<<i<<": "<<ans<<" Xukha"<<endl;
}
return 0;
}