【题解】FJUTOJ1309: Bi-shoe and Phi-shoe

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Problem 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.

SampleInput

3

5

1 2 3 4 5

6

10 11 12 13 14 15

2

1 1

SampleOutput

Case 1: 22 Xukha

Case 2: 88 Xukha

Case 3: 4 Xukha

?

题意：

T组数据，每组数据给定1<=n<=1e4，接下来n个整数1<=xi<=1e6，对于每一个xi，找出最小的yi，使euler(yi) >= xi。求yi之和。

思路：

由欧拉函数定义可知，euler(x) < x。所以对每一个x，找大于它的第一个质数即为答案。线性筛法打素数表，二分查找。

AC代码：92MS

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <cstdlib>
 5 #include <algorithm>
 6 using namespace std;
 7 typedef long long ll;
 8 
 9 const int MAXN = 1e6 + 15;
10 const int PRIME_MAX = MAXN / 2;
11 bool vis[MAXN];
12 int prime[PRIME_MAX],top;
13 
14 void init() {
15     ll i,j;
16     top = 0;
17     vis[1] = false;
18     for(i = 2; i < MAXN; ++i) {
19         if(!vis[i]) {
20             prime[top++] = i;
21         }
22         for(j = 0; prime[j] * i < MAXN; ++j) {
23             vis[prime[j] * i] = true;
24             if(i % prime[j] == 0)
25                 break;
26         }
27     }
28 }
29 
30 int main() {
31     init();
32     int t,n,i,now,k;
33     scanf("%d",&t);
34     for(k = 1; k <= t; ++k) {
35         ll ans = 0;
36         scanf("%d",&n);
37         for(i = 1; i <= n; i++) {
38             scanf("%d",&now);
39             ans += *upper_bound(prime,prime + top,now);
40         }
41         printf("Case %d: %lld Xukhan",k,ans);
42     }
43     return 0;
44 }