LightOJ-1370 Bi-shoe and Phi-shoe (欧拉函数+二分)

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

 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <cstdlib>
 #include <algorithm>
 using namespace std;
 typedef long long ll;

 const int MAXN = 1e6 + ;
 const int PRIME_MAX = MAXN / ;
 bool vis[MAXN];
 int prime[PRIME_MAX], top;

 void init() {
     ll i, j;
     top = ;
     vis[] = false;
     for(i = ; i < MAXN; ++i) {
         if(!vis[i]) {
             prime[top++] = i;
         }
         for(j = ; prime[j] * i < MAXN; ++j) {
             vis[prime[j] * i] = true;
             if(i % prime[j] == )
                 break;
         }
     }
 }

 int main() {
     init();
     int t, n, i, now, k;
     scanf("%d", &t);
     for(k = ; k <= t; ++k) {
         ll ans = ;
         scanf("%d", &n);
         for(i = ; i <= n; ++i) {
             scanf("%d", &now);
             ans += *upper_bound(prime, prime + top, now);
         }
         printf("Case %d: %lld Xukha\n", k, ans);
     }
     return ;
 }