Bi-shoe and Phi-shoe（欧拉函数/素筛）题解

Bi-shoe and Phi-shoe

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

题意：给你n个欧拉函数值,找出每一个欧拉函数值大于等于所给值的数，并且相加和最小

思路1：用筛法求1~N的欧拉函数，然后打表每个欧拉函数值的最优解，再取和最小

思路2：因为对于素数Φ（N）=N-1，所以给出p只要找出大于等于p+1的素数即可，用素筛

参考：很详细的欧拉函数解释

代码1：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<queue>
#include<cmath>
//#include<map>
#include<string>
#include<iostream>
#include<algorithm>
#define INF 0x3f3f3f3f
const int N=1000100;
const int MOD=1000;
using namespace std;
int euler[N];
int ans[N];
void init(){
	memset(ans,-1,sizeof(ans));
	for(int i=0;i<N;i++){
		euler[i]=i;
	}
	for(int i=2;i<N;i++){
		if(euler[i]==i){
			for(int j=i;j<N;j+=i){
				euler[j]=euler[j]/i*(i-1);	//f(n)=n*(1-1/p1)(1-1/p2)....(1-1/pk)
			}
		}
	}
	int now=0;    
	for(int i=2;i<N;i++){	//1不符合
		if(euler[i]>now && ans[euler[i]]==-1){
			ans[euler[i]]=i;
			now=euler[i];
		}
	}
}
int main(){
	int T,t,n;
	init();
	scanf("%d",&T);
	for(t=1;t<=T;t++){
		scanf("%d",&n);
		long long sum=0;
		while(n--){
			int p;
			scanf("%d",&p);
			for(int i=p;;i++){
				if(ans[i]!=-1){
					sum+=ans[i];
					break;
				}
			}
		}
		printf("Case %d: %lld Xukha\n",t,sum);
	}
	return 0;
}

代码2：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<queue>
#include<cmath>
//#include<map>
#include<string>
#include<iostream>
#include<algorithm>
#define INF 0x3f3f3f3f
const int N=1000100;
const int MOD=1000;
using namespace std;
int prime[N];
void init(){
	memset(prime,0,sizeof(prime));
	prime[0]=prime[1]=1;
	for(int i=2;i<N;i++){
		if(!prime[i]){
			for(int j=i*2;j<N;j+=i){
				prime[j]=1;
			}
		}
	}
}
int main(){
	int T,t,n;
	init();
	scanf("%d",&T);
	for(t=1;t<=T;t++){
		scanf("%d",&n);
		long long sum=0;
		while(n--){
			int p;
			scanf("%d",&p);
			p++;
			while(prime[p]!=0){
				p++;
			}
			sum+=p;
		}
		printf("Case %d: %lld Xukha\n",t,sum);
	}
	return 0;
}