Description
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
题意：
给一些数Ai（第 i 个数），Ai这些数代表的是某个数欧拉函数的值，我们要求出数 Ni 的欧拉函数值不小于Ai。而我们要求的就是这些 Ni 这些数字的和sum，而且我们想要sum最小，求出sum最小多少。
解题思路：
我们知道，一个素数P的欧拉函数值ψ(P)=P-1。所以如果我们知道ψ(N)，那么最小的N就是最接近N-1，并且大于ψ(N)的素数。我们把所有素数打表之后再判断就可以了。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 6 using namespace std; 7 const long long maxn=1010000; 8 bool b[maxn]; 9 long long p[maxn],k; 10 11 void is_prime() 12 { 13 memset(b,true,sizeof(b)); 14 memset(p,0,sizeof(p)); 15 b[0]=b[1]=false; 16 for(long long i=2;i<=maxn;i++) 17 { 18 if(b[i]) 19 { 20 p[k++]=i; 21 for(int j=i+i;j<=maxn;j+=i) b[j]=false; 22 } 23 } 24 } 25 26 int phi(long long tmp) 27 { 28 int l=0,r=k; 29 while(l<=r) 30 { 31 long long mid=(l+r)/2; 32 if(p[mid]>tmp) r=mid-1; 33 else l=mid+1; 34 } 35 for(int i=max(r,0);;i++) 36 if(p[i]>tmp) 37 return p[i]; 38 } 39 40 int main() 41 { 42 is_prime(); 43 44 long long n,m,l,sum=0; 45 cin>>n; 46 long long ans=1; 47 while(n--) 48 { 49 cin>>m; 50 while(m--) 51 { 52 cin>>l; 53 sum+=phi(l); 54 } 55 56 printf("Case %lld: %lld Xukha\n",ans,sum); 57 ans++; 58 sum=0; 59 } 60 61 return 0; 62 }
