【欧拉函数】Bi-shoe and Phi-shoe【LightOJ1370】------长篇阅读专场D题

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

解题思路：

这道题要用到欧拉函数的知识。

http://blog.csdn.net/sentimental_dog/article/details/52002608

但是我做的时候直接用素数筛法打表，然后输出大于等于输入的最小素数的和，就可以了。

这种做法是投机取巧的，正确做法还是应该用筛法求欧拉函数。

要注意，最后输出很大，必须用long long int型，否则会WA。

#include<stdio.h>
#include<iostream>
using namespace std;

int input[10010];
int prime[1001000]={0};
long long int ans;

void isprime()//用筛法将素数打表
{
prime[1]=1;
for(int i=2;i<=1001000;i++)
{
if(prime[i]==0)
{
for(int j=2*i;j<=1001000;j+=i)
{
prime[j]=1;
}
}
}
}

int main()
{
int t;
isprime();
scanf("%d",&t);
for(int i=1;i<=t;i++)
{
int n;
scanf("%d",&n);
for(int j=1;j<=n;j++)
{
scanf("%d",&input[j]);
}
ans=0;
for(int j=1;j<=n;j++)
{
int k=input[j]+1;
while(1)
{
if(prime[k]==0)
{
ans+=k;
break;
}
k++;
}
}
cout<<"Case "<<i<<": "<<ans<<" Xukha"<<endl;
}
}