题意

给出一些数字，对于每个数字找到一个欧拉函数值大于等于这个数的数，求找到的所有数的最小和。

Solution

线性筛出phi，把询问数组排序搞就行了

# include <bits/stdc++.h>

# define RG register

# define IL inline

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

typedef long long ll;

const int _(1e4 + 10), __(1e6 + 10);

IL ll Read(){

	char c = '%'; ll x = 0, z = 1;

	for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;

	for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';

	return x * z;

}

int n, a[_], id[_], prime[__], isprime[__], num, phi[__];

ll ans;

IL void Prepare(){

	for(RG int i = 2; i <= __; ++i){

		if(!isprime[i]) prime[++num] = i, phi[i] = i - 1;

		for(RG int j = 1; j <= num && i * prime[j] <= __; ++j){

			isprime[i * prime[j]] = 1;

			if(!(i % prime[j])){  phi[i * prime[j]] = phi[i] * prime[j]; break;  }

			else phi[i * prime[j]] = phi[i] * (prime[j] - 1);

		}

	}

}

int main(RG int argc, RG char *argv[]){

	Prepare();

	for(RG int T = Read(), i = 1; i <= T; ++i){

		n = Read(); ans = 0;

		for(RG int j = 1; j <= n; ++j) a[j] = Read();

		sort(a + 1, a + n + 1);

		for(RG int j = 1, l = 1; j <= n; ++j){

			while(phi[l] < a[j]) ++l;

			ans += l;

		}

		printf("Case %d: %lld Xukha\n", i, ans);

	}

	return 0;

}