[LightOJ](1370)Bi-shoe and Phi-shoe ---- 欧拉函数与素数筛（规律）

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个正整数，称为幸运数，让你找出n个>=ni的phi(x)，保证这找到的这n个x的值的和最小。
思路： 打一个欧拉函数的表，和幸运数字之间找规律。
发现满足题意的x都是质数p，又因为φ（p） = p-1，所以我们会发现，>=ni的phi(x) 中满足题意的就是那个最小的质数x。对于每一个幸运数ni我们直接从ni+1开始找，找到的第一个质数就是满足条件的x。

AC代码：

#include<bits/stdc++.h>
#define rep(i,s,e)      for(int i=s;i<=e;i++)
#define rev(i,s,e)      for(int i=e;i>=s;i--)
using namespace std;
const int maxn = 1e7+5;
typedef long long LL;
bool prime[maxn];
void Init()
{
    memset(prime,true,sizeof(prime));
    prime[0] = prime[1] = false;
    for(int i=2;i*i<maxn;i++)
    {
        if(prime[i])
            for(int j=i*i;j<maxn;j+=i) prime[j] = false;
    }
}
int main()
{
    #ifdef LOCAL
    freopen("in.txt","r",stdin);
    #endif // LOCAL
    Init();
    int t,n,e;
    int id=1;
    scanf("%d",&t);
    while(t--)
    {
        LL sum = 0;
        scanf("%d",&n);
        rep(i,1,n)
        {
            scanf("%d",&e);
            for(int j=e+1;;j++)
                if(prime[j])
                {
                    sum+=j;
                    break;
                }
        }
        printf("Case %d: %lld Xukha\n",id++,sum);
    }
    return 0;
}