LightOj 1370 Pi shoe and Phi shoe

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

此题利用到一个性质：素数p的欧拉函数为p-1，且两个素数之间的非素数的欧拉函数的值小于第一个素数的欧拉函数的值

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
bool prime[10000000];
int main(){
	int n,m,num;
	long long int re;
	memset(prime,false,sizeof(prime));
	for(int i=2;i<10000000;i++){
		if(prime[i]==false){
			for(int j=i+i;j<10000000;j+=i){
				prime[j]=true;
			}
		}
	}
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>m;
		re=0;
		for(int j=1;j<=m;j++){
			cin>>num;
			for(int t=num+1;;t++){
				if(prime[t]==false){
					re+=t;
					break;
				}
			}
		}
		printf("Case %d: %lld Xukha\n",i,re);
	}
	return 0;
}