LightOJ 1370 Bi-shoe and Phi-shoe (欧拉筛)

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 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].

For each case,print the case number and the minimum possible money spent for buying the bamboos. See the samples for details.

3

5

1 2 3 4 5

6

10 11 12 13 14 15

2

1 1

Case 1: 22 Xukha

Case 2: 88 Xukha

Case 3: 4 Xukha

题意：给你n个数a，然后求出最小的eular（i）>a，然后求和

思路：典型的欧拉筛，注意的是要从a+1开始寻找，因为有1的因素

总结：经典应用，不难

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,sizeof(x))
#define PI acos(-1)
using namespace std;
LL ans[MAXN];
void db()
{
	memset(ans,sizeof(ans));
	ans[1]=1;
	int i,j;
	for(i=2;i<=MAXN;i++)
	{
		if(!ans[i])
		{
			for(j=i;j<=MAXN;j+=i)
			{
				if(!ans[j])
				ans[j]=j;
				ans[j]=ans[j]/i*(i-1);
		    }
		}
	}
}
int main()
{
	int t,n,m,i,j,k;
	int cas=0;
	db();
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		ll sum=0;
		for(i=0;i<n;i++)
		{
			int a;
			scanf("%d",&a);
			int x=a+1;
			while(ans[x]<a)
			x++;
			printf("x=%dn",x);
			sum+=x;
		}
		printf("Case %d: %lld Xukhan",++cas,sum);
	}
	return 0;
}