【LightOJ】1370 - Bi-shoe and Phi-shoe（欧拉函数，素数打表）

1370 - Bi-shoe and Phi-shoe

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 thann which are relatively prime (having no common divisor other than 1) ton. 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 containsn 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.

这里用欧拉函数，即：一个素数的比它小的互质数的个数为该数减一。这道题之用了这一点。

但是这里还要注意一个知识点，假如ψ(N)的欧拉函数值为p，则N的最小值为大于p的最小素数。

代码如下：

#include <cstdio>
int su[1000111]={1,1};
int main()
{
	for (int i=2;i<=1000111;i++)		//打素数表，素数为0 
	{
		if (su[i]==1)
			continue;
		for (int j=i+i;j<=1000111;j+=i)
			su[j]=1;
	}
	int u;
	int n;
	long long ans;
	int num=1;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%d",&n);
		ans=0;
		while (n--)
		{
			int t;
			scanf ("%d",&t);
			for (int i=t+1;;i++)
			{
				if (su[i]==0)
				{
					ans+=i;
					break;
				}
			}
		}
		printf ("Case %d: %lld Xukha\n",num++,ans);
	}
	return 0;
}