题目链接：http://lightoj.com/login_main.php?url=volume_showproblem.php?problem=1370
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 nwhich 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：一个人教练要为他的学生们买竹竿，竹竿的价格=主干的长度，主干的价值=竹竿长度的欧拉值（质因子的个数），每个学生都有一个幸运数字，教练想要每个同学的竹竿的价值都大于等于他们的幸运数字，教练又想少花钱，输出满足条件的最少花费。
方法一：对于每个幸运数字，最近的就是下一个素数 （ 对于素数x： Φ（x）=x-1 ）
因此，打一个素数表，遍历求和即可。
方法二：直接打一个欧拉表，查找即可。
ac：
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<map>
//#include<set>
#include<deque>
#include<queue>
#include<stack>
#include<bitset>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define INF 0x3f3f3f3f
#define mod 1000000007
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b)
#define clean(a,b) memset(a,b,sizeof(a))// 水印
//std::ios::sync_with_stdio(false);
ll oula[1000100],arr[1000100];
/*ll euler(ll n)
{
ll res=n;
for(ll i=2;i*i<=n;++i)
{
if(n%i==0)
{
res=res/i*(i-1);
while(n%i==0)
n=n/i;
}
}
if(n>1)
res=res/n*(n-1);
return res;
}*/
void intt()
{
clean(oula,0);
clean(arr,0);
oula[1]=1;
for(int i=2;i<1000100;++i)
{
if(oula[i]==0)
{
for(int j=i;j<1000100;j=j+i)
{
if(oula[j]==0)
oula[j]=j;
oula[j]=oula[j]-oula[j]/i;
}
}
}
}
int main()
{
intt();
int t,num=1;
cin>>t;
while(t--)
{
int n;
cin>>n;
ll ans=0;
for(int i=0;i<n;++i)
{
int x;
cin>>x;
for(int j=x+1;j<1000100;++j)
{
if(oula[j]>=x)//符合要求的欧拉值
{
ans=ans+j;
break;
}
}
}
cout<<"Case "<<num++<<": "<<ans<<" Xukha"<<endl;
}
}