LightOJ 1370- Bi-shoe and Phi-shoe-白红宇

LightOJ 1370- Bi-shoe and Phi-shoe

阅读量：5325 次

发布时间：2019-06-14

本文共 2830 字，大约阅读时间需要 9 分钟。

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

1 2 3 4 5

10 11 12 13 14 15

1 1

Sample Output

Case 1: 22 Xukha

Case 2: 88 Xukha

Case 3: 4 Xukha

Score就是对应长度的欧拉函数

从lucky number+1开始递增选择第一个出现的质数作为cost

代码

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 #include 
        
          5 #include 
         
           6 #include 
          
            7 #include 
           
             8 #include 
            
              9 #include 
             
              10 #include 
              
               11 #include 
               
                12 #include 
                
                 13 #include 
                 
                  14 #include 
                  
                   15 #include 
                   16 #include 
                    
                     17 #include 
                     
                      18 #include 
                      
                       19 #include 
                       
                        20 #include 
                        
                         21 #include 
                         
                          22 #include 
                          
                           23 #include 
                           
                            24 #include 
                            
                             25 26 #define gc getchar()27 #define mem(a) memset(a,0,sizeof(a))28 #define mod 100000000729 #define sort(a,n,int) sort(a,a+n,less
                             
                              ())30 #define fread() freopen("in.in","r",stdin)31 #define fwrite() freopen("out.out","w",stdout)32 using namespace std;33 34 typedef long long ll;35 typedef char ch;36 typedef double db;37 38 const int maxn=1e5+10;39 int aa[maxn];40 41 int fun(int a)42 {43 int i , m;44 for(int n = a+1;;n++)45 {46 m = sqrt(n);47 for(i=2;i<=m;i++)48 {49 if(n%i==0) break;50 }51 if (i>m)52 {53 return n;54 } 55 }56 } 57 58 int main()59 {60 int t , n , sum = 0;61 int a[10005] = { 0};62 cin >> t;63 for(int j = 1;j<=n;j++)64 {65 cin >> n;66 for(int i = 0;i
                              
                               >a[i];69 }70 for(int i = 0;i

转载于:https://www.cnblogs.com/SutsuharaYuki/p/11230283.html

你可能感兴趣的文章