A - Bi-shoe and Phi-shoe
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:t个测试样例,每个样例 n 个数,对于给定的每个数a[i],要求一个数x,满足x的欧拉函数值大于a[i]
要找n个满足条件的数x,并且要求n 个数x 的和 cnt 最小,输出 cnt
题解:一个素数 p 的欧拉函数值等于 p-1;所以要找满足条件:欧拉函数值大于a[i] 的最小x;只要找大于a[i] 的最小素数即可,最后输出n个素数的和
#include<iostream> #include<algorithm> #include<math.h> #include<string.h> #include<string> #define ll long long #define mx 1000010 using namespace std; int prim[mx]; void init()//素数打表 { memset(prim,0,sizeof(prim)); prim[1]=1; for(int i=2;i<mx;i++) { if(prim[i])//是偶数 continue; for(int j=i<<1;j<mx;j=j+i)//把素数的倍数标记 prim[j]=1; } } int main() { int t,n,x; cin>>t; init(); for(int i=1;i<=t;i++) { cin>>n; ll cnt=0; for(int j=0;j<n;j++) { cin>>x; for(int k=x+1;k<mx;k++) { if(prim[k]==0) { cnt=cnt+k; break; } } } cout<<"Case "<<i<<": "<<cnt<<" Xukha"<<endl; } return 0; }