lightoj1370欧拉函数

Posted 2020-10-03 不积跬步无以至千里，不积小流无以成江海

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

bamboo的score值就是其长度x的欧拉函数值（即小于x且与x互质的数的个数）欧拉函数有一个性质：素数p的欧拉函数值为p-1；

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 int a[1000000+10];
 5 int t,n,x;
 6 long long sum;
 7 int b[1000000+10]={1,1,0};
 8 void f()
 9 {
10     for(int i=2;i<=1000000+9;i++)
11     {
12         if(!b[i])
13         {
14             for(int j=i+i;j<=1000000+9;j+=i)
15                 b[j]=1;
16         }
17     }
18 }
19 int main()
20 {
21     f();
22     while(scanf("%d",&t)!=EOF)
23     {
24         for(int w=1;w<=t;w++)
25         {
26             sum=0;
27             scanf("%d",&n);
28             for(int i=1;i<=n;i++)
29             {
30                 scanf("%d",&x);
31                 for(int k=x+1;k<=1000000+9;k++)
32                 {
33                     if(b[k]==0)
34                     {
35                         sum+=k;
36                         break;
37                     }
38                 }
40             }
41             printf("Case %d: %lld Xukha\n",w,sum);
42         }
43 
44 
45     }
46     return 0;
47 }