lightoj 1370欧拉函数

1370 - Bi-shoe and Phi-shoe

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
3 5 1 2 3 4 5 6 扫描二维码关注公众号，回复： 1493550 查看本文章 10 11 12 13 14 15 2 1 1	Case 1: 22 Xukha Case 2: 88 Xukha Case 3: 4 Xukha

题意：每一个数字的欧拉函数要大于或等于该数字。求，最小的欧拉函数的下标和的大小。

答案要用longlong存。

 1 #include<cstdio>
 2 using namespace std;
 3 const int maxn=1000006;
 4 int prime[maxn];
 5 
 6 void init()
 7 {
 8     for(int i=1;i<maxn;i++)
 9         prime[i]=i;
10     for(int i=2;i<maxn;i++){
11         if(prime[i]==i){
12             int temp=i;
13             for(int j=i;j<maxn;j+=temp){
14                 prime[j]=prime[j]/temp*(temp-1);
15             }
16         }
17     }
18 }
19 int main()
20 {
21     int n,T;
22     init();
23     scanf("%d",&T);
24     for(int kkk=1;kkk<=T;kkk++){
25         scanf("%d",&n);
26         long long ans=0;
27         while(n--){
28             int x,temp;
29             scanf("%d",&x);
30             temp=x+1;
31             while(prime[temp]<x)
32                 temp++;
33             ans+=temp;
34         }
35         printf("Case %d: %lld Xukha\n",kkk,ans);
36     }
37 
38     return 0;
39 }