LightOJ

Bi-shoe and Phi-shoe

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

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Description

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

Source

Problem Setter: Mir Wasi Ahmed

Special Thanks: F.A. Rezaur Rahman Chowdhury, Jane Alam Jan

/**
          题意：f(n) 表示 小于等于 n 的数中素数的个数；给出一串数 比如x 求f(x) >= x 的最小和
          做法：欧拉函数 
**/
#include <iostream>
#include<cmath>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<stack>
#define maxn 1000000 + 10
int mindiv[maxn],phi[maxn],sum[maxn];
int mmap[maxn];
int mmpp[maxn];
using namespace std;
void solve()
{
    for(int i=1; i<maxn; i++)
    {
        mindiv[i] = i;
    }
    for(int i=2; i*i<maxn; i++)
    {
        if(mindiv[i] == i)
        {
            for(int j=i*i; j<maxn; j+=i)
            {
                mindiv[j] = i;
            }
        }
    }
    phi[1] = 1;
    for(int i=2; i<maxn; i++)
    {
        phi[i] = phi[i/mindiv[i]];
        if((i/mindiv[i])%mindiv[i] == 0)
        {
            phi[i] *=mindiv[i];
        }
        else
        {
            phi[i] *= mindiv[i] -1;
        }
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
          int T;
          scanf("%d",&T);
    int n;
    solve();
    int Case = 1;
    while(T--)
    {
                    scanf("%d",&n);
        int res = 0;
        long long sum = 0;
        for(int i=0; i<n; i++)
        {
            scanf("%d",&mmpp[i]);
        }
        sort(mmpp,mmpp+n);
        int tt=  0,Index = 0;
        int j = 2;
        phi[1] = 1;
        for(int i=0; i<n; )
        {
            if(phi[j] >= mmpp[i])
            {
                sum += j;
                i++;
            }
            else j++;
        }
        printf("Case %d: %lld Xukha
",Case++,sum);
    }
    return 0;
}