数论-欧拉函数-Bi-shoe and Phi-shoe LightOJ - 1370

          Bi-shoe and Phi-shoe LightOJ - 1370    

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

题目大意：

一个bamboo的分 = Φ (l)；
其中 l 为bamboo的长度，Φ (n)为小于n的正整数（除 1 以外）的相对素数的个数（即$n$不是$i$的倍数）;
给定n个数，求大于等于Φ (i)的和；

思路：

最朴素的做法是根据定义找每个数的价值，然后打成一张表，然后一个个去比对，但是那样会超时。

首先搞懂欧拉函数是什么？？？
欧拉函数指的是 n以内与n互质的所有数的个数。

我们知道欧拉函数有一个性质：素数p的欧拉函数值为p-1；
我们要找长度为小的bamboo，只需要找x+1以后的第一个素数。

实现代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>

using namespace std;
#define ll long long

int t,n;
bool f[1000007];
ll s = 0;

int main()
{
    memset(f,0,sizeof(f));
    for(int i = 2; i < 1007; i++) if(!f[i])
        for(int j = i*i; j < 1000007; j+=i) f[j] = true;
    scanf("%d",&t);
    int caset = 1,x;
    while(t--)
    {
        s = 0;
        scanf("%d",&n);
        for(int i = 0; i < n; i++)
        {
            scanf("%d",&x);
            for(int j = x + 1; j < 1000007; j++)
            if(!f[j]){
                s += j;
                break;
            }
        }
        printf("Case %d: %lld Xukha\n",caset++,s);
    }
    return 0;
}