Bi-shoe and Phi-shoe 线筛欧拉函数

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

首先最重要的 要读懂这题让干嘛 = =
就是给一串数 这些数是某个欧拉函数的值 找到最小自变量并求和
即phi（n） = x； x已知 求满足phi（n） = x 的n的最小值 并对所有xi求和

#include <cstdio>#include <algorithm>using namespace std;#define ll long longconst int N = 1000000+5;int phi[N];int num[10005];// 线筛欧拉函数void euler(){    phi[1] = 1;    for (int i = 2; i < N; ++i){        if (!phi[i]){            for (int j = i; j < N; j += i){                if (!phi[j]) phi[j] = j;                phi[j] = phi[j]/i*(i-1);            }        }    }}int main(){    euler();    int t;    int kase = 1;    scanf("%d",&t);    while (t--){        int n;        scanf("%d",&n);        for (int i = 0; i < n; ++i){            scanf("%d",&num[i]);        }        sort(num,num+n);        ll sum = 0; //注意是 long long 10^6*10000        for (int i = 0, j = 2;i < n;){            if (phi[j] >= num[i]){                sum+=j;                i++;            }else j++;        }        printf("Case %d: %lld Xukha\n",kase++,sum);    }    return 0;}