本文共 2343 字，大约阅读时间需要 7 分钟。

题目：

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

代码如下：

#include 
   
    #include 
    
     #include 
     
      #include 
      
       using namespace std;const int maxn=1e6+1e5;typedef long long ll;int pri[maxn];int ou[maxn];int t;int n;int a[maxn];int ha[maxn];void qiu (){    memset (pri,0,sizeof(pri));    for (int i=1;i
       
        =0&&ha[j]==0;j--)        {            ha[j]=i;        }    }}int main(){    memset (ha,0,sizeof(ha));    qiu();    scanf("%d",&t);    for (int j=1;j<=t;j++)    {        scanf("%d",&n);        for (int i=1;i<=n;i++) scanf("%d",&a[i]);        ll sum=0;        for (int i=1;i<=n;i++)        {            //printf("ha[%d]=%d\n",a[i],ha[a[i]]);            sum+=ha[a[i]];        }        printf("Case %d: %lld Xukha\n",j,sum);    }    return 0;}
       
      
     
    
   

转载地址：http://jkoen.baihongyu.com/