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
如果每个数都做为欧拉函数的值的话，那这个数最小就是大于这个值的素数；
例如：1～2，2～3，3～5，4～5，5～7，6～7……
故想到用素数打表就可以实现：
主要思路：素数筛选
先设置a[MAXN]做为标记数组；先将a[1]标记为1,从2开始，如果2不为1，将2的倍数标为1，
如此循环：
2，3，4，5，6，7，8，9，10，11，12，13，14，15，16，17，18，19，20…….
2，3，5，7，9，11，13，15，17，19，，，，，，
2，3，5，7，11，13，19，，，，，，
最后筛选下来的都是素数；可以在其中将素数储存在另个数组（打表）**
最后注意，%I64d 与%lld的使用，该用%lld时用另一个会直接导致代码出错
代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
const ll MAXN = 1000005;
ll a[MAXN],b[MAXN];
ll t,n,sum;
int T;
void su()
{
memset(a,0,sizeof(a));
a[1] = 1,t=1;
for(ll i=2;i<MAXN;i++)
{
if(!a[i]){//每个素数作为头
while(t<i) //打表
{
b[t] = i;
t++;
}
for(ll j=2*i;j<MAXN;j+=i)//一直查找素数头的倍数并标记
{
if(!a[j])
a[j] = 1;
}
}
}
}
int main()
{
su();
scanf("%d",&T);
for(int i=1;i<=T;i++)
{
scanf("%lld",&n);
sum = 0;
for(ll j=1;j<=n;j++)
{
int z;
scanf("%d",&z);
sum += b[z];
}
printf("Case %d: %lld Xukha\n",i,sum);
}
return 0;
}