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：求n个后面的每一个数的第一个素数并求和。
比较笨，先打表把素数标出来再取出来，再把每一位数后面的素数打出来。就这么胡来…
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <cmath> 5 6 using namespace std; 7 8 typedef long long LL; 9 const int maxn = 1000010; 10 bool isprime[maxn]; 11 LL p[maxn]; 12 LL m[maxn]; 13 14 void printlist() { 15 memset(isprime, true, sizeof(isprime)); 16 isprime[0] = isprime[1] = false; 17 int pedge = int(sqrt(maxn)); 18 for(int i = 2; i <= pedge; i++) { 19 if(isprime[i]) { 20 int o = maxn / i; 21 for(int j = 2; j <= o; j++) { 22 isprime[i*j] = false; 23 } 24 } 25 } 26 } 27 28 void init() { 29 printlist(); 30 int k = 0; 31 for(int i = 0; i < maxn; i++) { 32 if(isprime[i]) { 33 p[k++] = i; 34 } 35 } 36 k = 0; 37 for(int i = 0; i < maxn; i++) { 38 if(i < p[k]){ 39 m[i] = p[k]; 40 continue; 41 } 42 m[i] = p[++k]; 43 } 44 } 45 46 int main() { 47 int T, n; 48 init(); 49 scanf("%d", &T); 50 for(int ii = 1; ii <= T; ii++) { 51 scanf("%d", &n); 52 LL ans = 0; 53 int tmp; 54 while(n--) { 55 scanf("%d", &tmp); 56 ans += m[tmp]; 57 } 58 printf("Case %d: %lld Xukha ", ii, ans); 59 } 60 }
Hint