LightOJ1370 Bi-shoe and Phi-shoe —— 欧拉函数

题目链接：https://vjudge.net/problem/LightOJ-1370

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 10 11 12 13 14 15 2 1 1	Case 1: 22 Xukha Case 2: 88 Xukha Case 3: 4 Xukha

题意：

给出n个数，为每个数x找到满足：x<=Euler(y) 的最小的y，其中Euler()为欧拉函数。

题解：

可知，当y为素数，Euler(y) = y-1。所以，只需从x+1开始，找到第一个素数即可。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = 1e6+;

 bool vis[MAXN];

 void getPrime()
 {
     int m = (int)sqrt(MAXN);
     memset(vis, , sizeof(vis));
     vis[] = ;
     for(int i = ; i<=m; i++) if(!vis[i])
         for(int j = i*i; j<MAXN; j+=i)
             vis[j] = ;
 }

 int main()
 {
     getPrime();
     int T, n;
     scanf("%d", &T);
     for(int kase = ; kase<=T; kase++)
     {
         LL ans = ;
         scanf("%d", &n);
         for(int i = ; i<=n; i++)
         {
             int x;
             scanf("%d", &x);
             for(int j = x+;;j++) if(!vis[j]){
                 ans += j;
                 break;
             }
         }
         printf("Case %d: %lld Xukha\n", kase,ans);
     }
 }