Bi-shoe and Phi-shoe(欧拉函数)

最新推荐文章于 2025-03-04 18:06:41 发布

原创最新推荐文章于 2025-03-04 18:06:41 发布 · 2.9k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#欧拉函数 #素数

数学+DP 专栏收录该内容

116 篇文章

订阅专栏

本篇介绍了一个关于竹竿跳高的趣味算法问题，通过购买不同长度的竹竿来满足学生们的幸运数字需求，利用欧拉函数计算竹竿得分，并采用质数筛选法寻找最优解。

A - Bi-shoe and Phi-shoe

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status

Description

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

题目链接：http://www.bnuoj.com/bnuoj/problem_show.php?pid=13288

被这个题目最后的输出格式气得半天，__int64输出用lld输出才AC，用I64d不行。

欧拉函数定义：比他小的并与他互质的个数之和。

题意给你n个luky数字是欧拉函数的值。也就是个数，求和最小，我就想应该是比这个数大的质数就肯定是最小的，这样先把所有质数枚举出来就好办了。用到了素数筛选法。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>
#include <ctime>
#define LL __int64
#define eps 1e-8
#define pi acos(-1)
using namespace std;
int a[1010010];
int p[100000];
void INIT(){
	memset(a,0,sizeof(a));
	memset(p,0,sizeof(p));
	int t=0;
	for (int i=2;i<=1010000;i++){
		if (a[i]==0)
			p[++t]=i;
		for (int j=1,k;j<=t && ((k=i*p[j])<=1010000);j++)
		{
			a[k]=1;
			if (i % p[j]==0) break; 
		} 
	}
	t=1;
	for (int i=1;i<=1000001;i++){
		if (i<p[t]) a[i]=p[t];
		else {
			t++;
			a[i]=p[t];
		}
	}
}
int main(){
	INIT();
	int T;
	//freopen("in.txt","r",stdin);
	//fropen("out.txt","w",stdout)
	scanf("%d",&T);
	for (int cas=1;cas<=T;cas++){
		int n,k;
		scanf("%d",&n);
		LL ans=0;
		for(int i=1;i<=n;i++){
			scanf("%d",&k);
			ans+=a[k];
		}
		printf("Case %d: %lld Xukha\n",cas,ans);
	}
	return 0;
}