Bi-shoe and Phi-shoe_k什么of bamboos-优快云博客

博客围绕竹杆购买问题展开，教练需为学生买竹杆，竹杆得分是其长度的欧拉函数值，学生有幸运数字，要使竹杆得分大于等于幸运数字且花费最少。输入多组数据，每组含学生数及幸运数字，题解运用素数筛打表找最小解。

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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

1 2 3 4 5

10 11 12 13 14 15

1 1

Sample Output

Case 1: 22 Xukha

Case 2: 88 Xukha

Case 3: 4 Xukha

题解：输入t组数据，每组输入n，表示接下来n个数ai，每个ai表示某个数（假设xi）的欧拉函数值，让求所有xi的最小和。注意每个ai对应无数个xi，找出最小的相加就行。这里运用素数筛先进行打表，将所有素数标记，这样每次输入ai可以尽快找出最小xi。

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