[LightOJ1370]Bi-shoe and Phi-shoe

A - Bi-shoe and Phi-shoe
Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu
 

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\n", ii, ans);
59     }
60 }

 

posted @ 2015-08-19 18:57  Kirai  阅读(210)  评论(0)  编辑  收藏  举报