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
每个人有一个幸运值n，求欧拉函数Φ(x)>=n的最小的那个x。
根据欧拉函数的定义，素数n的欧拉函数值一定是n-1，所以在两个素数的区间(n1,n2)中的数（这肯定是合数）的欧拉函数值肯定小于Φ(n2)。所以给定一个数x，距这个数x+1的最近的素数就是要求的数，因为x+1以下的数的欧拉函数值肯定不会超过x。
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <map>
#include <cmath>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MAXN 1000005
#define mod 1000000007
using namespace std;
int p[MAXN];
void Init()
{
for(int i=2; i<MAXN; ++i)
{
if(!p[i])
{
for(int j=i+i; j<=MAXN; j+=i)
p[j]=1;
}
}
}
int main()
{
Init();
int t,cnt=1,n,m;
scanf("%d",&t);
while(t--)
{
long long ans=0;
scanf("%d",&n);
while(n--)
{
scanf("%d",&m);
for(int i=m+1;; ++i)
if(p[i]==0)
{
ans+=i;
break;
}
}
printf("Case %d: %lld Xukha\n",cnt++,ans);
}
return 0;
}