A-Bi-shoe and Phi-shoe LightOJ-1370 ------------------------------------- Number Theory (Linear Euler sieve)

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

Analysis:
Find the sum of the prime factors of each number [1,10000000]. Solving with a linear Euler sieve
must pay attention to sorting the array.
Then traverse the array to solve

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=1e6+10000;
int a[N];
ll phi[N],cnt;
int prime[N];
bool st[N];
int t,n;
void init()
{
	phi[1]=1;
	for(int i=2;i<=N;i++)
	{
		if(!st[i]) 
		{
			prime[++cnt]=i;
			phi[i]=i-1;
		}
		for(int j=1;j<=cnt&&prime[j]<=N/i;j++)
		{
			st[prime[j]*i]=true;
			if(i%prime[j]==0)
			{
				phi[i*prime[j]]=phi[i]*prime[j];
				break;
			}
			phi[i*prime[j]]=phi[i]*(prime[j]-1);
		}
	}
}
int main()
{
	init();
	int k=0;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(int i=1;i<=n;i++) scanf("%d",&a[i]);
		ll sum=0;
		sort(a+1,a+1+n);
		for(int i=1,j=1;i<=N&&j<=n;)
		{
			if(phi[i]>=a[j])
			{
				sum+=i;
				j++;
			}
			else i++;
		}
		printf("Case %d: %lld Xukha\n",++k,sum);
	 } 
 }