LightOJ1370 Bi-shoe and Phi-shoe

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;
}