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 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 nspace 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

题目大意：一个竹竿长度为p,它的score值就是比p长度小且与且与p互质的数字总数，比如9有1，2，4，5，7，8这六个数那它的score就是6。

每个学生都有一个幸运数字，要求买回来的竹子的score值能>=学生的幸运数字，每个竹子的长度就是花费，求最小花费。

思路：欧拉函数的性质：如果一个数是质数（n），那它的欧拉函数值为n-1。所以只要求出大于每个幸运数字的第一个素数。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll long long
int a[10010];
int su[1000010];
void prime()
{
    memset(su,0,sizeof(su));
    su[2]=0;//*********
    int i,j;
    for(i=2; i*i<=1000100; i++)
    {
        for(j=i*2; j<=1000100; j+=i)
            su[j]=1;
    }
}
int main()
{
    int t;
    prime();
    int o=1;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        memset(a,0,sizeof(a));
        scanf("%d",&n);
        for(int i=0; i<n; i++)
            scanf("%d",&a[i]);
        ll ans=0;
        for(int i=0; i<n; i++)
        {
            for(int j=a[i]+1;; j++)
            {
                if(!su[j])
                {
                    ans+=j;
                    break;
                }
            }
        }
        printf("Case %d: %lld Xukha\n",o++,ans);
    }
    return 0;
}