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
题目大意：就是最小的欧拉函数值；欧拉函数值就是小于或等于n的正整数中与n互质的数的数目；
解题思路：我们知道一个素数P的欧拉函数值ψ(P)=P-1。所以如果我们知道ψ(N)，那么最小的N就是最接近ψ(N)，
并且大于ψ(N)的素数。然后我们用欧拉筛打表算出最小值即可；
代码：
#include<stdio.h>
int a[1010011]= {0};
int b[1010011]= {0};
int main()
{
int t=0,m;
a[0]=1;
a[1]=1;
int i,j;
for(i=2; i<=1010100; i++) //用欧拉筛打素数表
{
if(!a[i]) b[t++]=i;
for(j=0; j<t; j++)
{
if(i*b[j]>1010101)
break;
a[i*b[j]]=1;
if(i%b[j]==0)
break;
}
}
t=0;
for(i=1; i<=1000001; i++) //
{
if(i<b[t]) a[i]=b[t]; //比素数小就是最小的素数
else
{
t++;
a[i]=b[t]; //大于这个数的最小素数
}
}
int n,k=1;
scanf("%d",&n);
while(n--)
{
int m;
scanf("%d",&m);
long long int sum=0;
int d;
for(i=0; i<m; i++)
{
scanf("%d",&d); //直接加上打好的表就行了
sum+=a[d];
}
printf("Case %d: %lld Xukha\n",k++,sum);
}
return 0;
}