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

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

代码:

欧拉筛

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<cmath>
#include<stdlib.h>

const int maxn=1e6+5;
typedef long long ll;
using namespace std;

int prime[1000005];
bool vis[1000005];
void erla() {
	int cnt =0;
	memset(vis,false,sizeof(vis));
	memset(prime,0,sizeof(prime));
	for(int t=2; t<=1000005; t++) {
		if(!vis[t]) {
			prime[cnt++]=t;
		}
		for(int j=0; j<cnt&&t*prime[j]<=1000005; j++) {
			vis[t*prime[j]]=true;
			if(t%prime[j]==0) {
				break;
			}
		}
	}
}

int main()
{
	int T;
    cin>>T;
    erla();
    int n;
    int cnt=1;
    while(T--)
    {
    	cin>>n;
    	ll sum=0;
    	int x;
    	int l,r;
    	for(int t=0;t<n;t++)
    	{
    		scanf("%d",&x);
    	    for(int j=x+1;j<=1000003;j++)
    	    {
    	    	if(vis[j]==false)
    	    	{
    	    		sum+=j;
    	    		break;
    	    	}
    	    }

    	}
    	cout<<"Case "<<cnt<<": "<<sum<<" Xukha"<<endl;
    	cnt++;
    }
	return 0;
}

二分+欧拉函数

wa了，没找到错

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<cmath>

const int maxn=1e6+5;
typedef long long ll;
using namespace std;

struct node
{
	ll val,id;
}p[maxn];
int Eular(int n)
{
	int ans = n;
	for (int i = 2 ; i * i <= n ; i++)
	{
		if (n % i == 0)
		{
			ans -= ans / i;
			while (n % i == 0)
				n /= i;
		}
	}
	if (n > 1)
		ans -= ans / n;
	return ans;
}
bool cmp(node x,node y)
{
	if(x.val!=y.val)
	{
		return x.val<y.val;
	}
	else
	{
		return x.id<y.id;
	}
}

int main()
{
int cc=0;
for(int t=2;t<=1000003;t++)
{
	p[cc].val=Eular(t);
	p[cc++].id=t;
}
sort(p,p+cc,cmp);
    int T;
    cin>>T;
    int n;
    int cnt=1;
    while(T--)
    {
    	cin>>n;
    	ll sum=0;
    	int x;
    	int l,r;
    	for(int t=0;t<n;t++)
    	{
    		scanf("%d",&x);
    		l=0;r=cc;
    		while(l<r)
    		{
    		    int mid=(l+r)>>1;
    		    if(p[mid].val<x)
    		    {
    		    	l=mid+1;
    		    }
    		    else
    		    {
    		    	r=mid;
    		    }
    		}
    		sum+=p[(l+r)>>1].id;

    	}
    	cout<<"Case "<<cnt<<": "<<sum<<" Xukha"<<endl;
    	cnt++;
    }

	return 0;
}