LightOJ - 1370【欧拉筛法*经典】

最新推荐文章于 2019-08-17 19:19:00 发布

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本文介绍了一个关于竹竿跳高的趣味数学问题，通过使用欧拉函数来解决如何为每位学生找到合适的竹竿长度，同时使得总成本最低。文章详细解释了问题背景、输入输出格式，并给出了两种解决方案：一种基于素数筛选，另一种则采用了更为复杂的预处理方法。

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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

1 2 3 4 5

10 11 12 13 14 15

1 1

Sample Output

Case 1: 22 Xukha

Case 2: 88 Xukha

Case 3: 4 Xukha

题意：给你一组数字，让你求欧拉函数的值大于等于它的数的和的最小值；

思路：第一种方法不难，重点是第二种方法的学习；

第一种：素数的euler应该是相同euler值最小值，只需要筛查素数就可已，euler（n）=n-1；

第二种，素数筛法预处理，用a数组替换后面euler值比较小的，用b数组表示euler值与最小下标的一个映射；

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int max_n=2000010;
int a[max_n+5],b[max_n+5];

void Init()//euler筛法 
{
	a[1]=0;
    for(int i=2;i<=max_n;i++)
    	a[i]=i;
	for(int i=2;i<max_n;i++)
	{
		if(a[i]==i)
		{
			for(int j=i;j<max_n;j+=i)
				a[j]=a[j]/i*(i-1);
		}
	}
}

int main()
{
	int T,n,k;
	long long ans;
    Init();
	memset(b,0,sizeof(b));
	for(int i=1;i<=max_n;i++) a[i]=max(a[i],a[i-1]);//不可省略 
	for(int i=max_n;i>=1;i--) b[a[i]]=i;
	for(int i=1;i<=max_n;i++)
	{
		if(b[i]==0)
		{
			int t=i;
			while(b[i]==b[t] && t<=max_n)
			{
				t++;
			}
			b[i]=b[t];
		}
	}
	b[0]=0;
	scanf("%d",&T);
	while(T--)
	{
		static int cnt=0;
		ans=0;
		scanf("%d",&n);
		while(n--)
		{
			scanf("%d",&k);
			ans+=b[k];
		}
		printf("Case %d: %lld Xukha\n",++cnt,ans);
	}
    return 0;
}