BNU 13288 Bi-shoe and Phi-shoe 【素数筛选】

A - 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

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <ctype.h>
#include <time.h>
#include <queue>
#include <iterator>

using namespace std;

const int MAXN = 1000100;
int n, m, t;
int prime[MAXN], ph[MAXN], p[MAXN], q[MAXN];

void solve(int n)
{
	memset(prime,0,sizeof(prime));
	memset(p,0,sizeof(p));
	memset(ph,0,sizeof(ph));
	int t = 0;

	for (int i = 2; i <= n; i++)
	{
		if (p[i] == 0)
			prime[++t] = i;
		for (int j = i * 2; j <= n; j+=i)
		{
			p[j] = 1;
		}
	}
	t = 1;
	for (int i = 1; i <= n; i++)
	{
		while (i >= prime[t])
		{
			t++;
		}
		if (i < prime[t])
			ph[i] = prime[t];
	}
}
/*
int phi[MAXN];

void Phi(int n)
{
	for (int i = 0; i <= n; i++)
		phi[i] = 0;
	phi[1] = 1;
	for (int i = 2; i <= n; i++)
	{
		if (!phi[i])
		{
			for (int j = i; j <= n; j += i)
			{
				if (!phi[j]) phi[j] = j;
				phi[j] = phi[j] / i*(i-1);
			}
		}
	}
}
*/
int main()
{
	int cases = 1;
	solve(1001000);
	scanf("%d",&t);
	while (t--)
	{
		long long ans = 0;
		scanf("%d",&n);
		for (int i = 0; i < n; i++)
			scanf("%d", &q[i]);

		sort(q,q+n);

		for (int i = 0; i < n; i++)
			ans += ph[q[i]];
		printf("Case %d: %lld Xukha\n",cases++,ans);
	}
	return 0;
}