AYIT-2020 609第十六周周赛 C - 素数筛

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在虚拟国度Xzhiland，竹竿跳高是一项热门运动，教练Master Phi-shoe需为学生选购竹竿，每根竹竿的分数由其长度的欧拉函数决定。助手Bi-shoe的任务是在满足每位学生幸运数字对应的最低分数要求下，以最少的花费完成采购。此问题转化为数学算法挑战，需找到一组数，使得这些数的欧拉函数值大于等于给定的幸运数字，同时这组数的总和最小。

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

给你 n 个数，分别找出大于等于这些数的欧拉函数值的 x ,使 x 的和最小

方法一：
进行素数打表，然后从x+1开始寻找并求和

#include <stdio.h>
int f[1000010]={1,1,0};
int main()
{
	for(int i=2;i*i<=1000010;i++)//素数筛
	{
		if(!f[i])
			for(int j=i*2;j<=1000010;j+=i)
				f[j]=1;
	}
	int t;
	scanf("%d",&t);
	int w=1;
	while(t--)
	{
		int n,x;
		long long s=0;
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		{
			scanf("%d",&x);
			for(int j=x+1;;j++)
			{
				if(!f[j])
				{
					s+=j;
					break;
				}
			}
		}
		printf("Case %d: %lld Xukha\n",w++,s);
	}
}

方法二：
这个题中的规定和欧拉函数不一样欧拉函数是小于等于 n , 所以在本题中 1 的值不为 1 而是 0

#include <iostream>
#include <algorithm>
#include <cstdio>
#define ll long long
using namespace std;
#define max 1123456

ll eular[max];

void init()
{
    eular[1]=1;
    for(int i=2;i<max;i++)
     eular[i]=i;
    for(int i=2;i<max;i++)
    {
        if(eular[i]==i)
        {
            for(int j=i;j<max;j+=i)
            eular[j]=eular[j]/i*(i-1);
        }
    }
}

int main()
{
    ll c,data[10005],ans,tmp=0;
    cin>>c;
    init();
    while(c--)
    {
        int n;
        tmp++;
        cin>>n;
        ans=0;
        for(int i=0;i<n;i++)
        cin>>data[i];
        for(int i=0;i<n;i++)
        {
            for(int j=data[i]+1;j<max;j++)
            if(eular[j]>=data[i])
            {
                ans+=j;
                break;
            }
        }
        printf("Case %lld: %lld Xukha\n",tmp,ans);
    }
    return 0;
}