Bi-shoe and Phi-shoe(欧拉函数)

Bi-shoe and Phi-shoe

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

Submit Status Practice LightOJ 1370

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

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

题意：

 

给一些数Ai（第 i 个数），Ai这些数代表的是某个数欧拉函数的值，我们要求出数 Ni 的欧拉函数值不小于Ai。而我们要求的就是这些 Ni 这些数字的和sum，而且我们想要sum最小，求出sum最小多少。

解题思路：

我们知道，一个素数P的欧拉函数值ψ(P)=P-1。所以如果我们知道ψ(N)，那么最小的N就是最接近N-1，并且大于ψ(N)的素数。我们把所有素数打表之后再判断就可以了。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const long long maxn=1010000;
bool b[maxn];
long long p[maxn],k;

void is_prime()
{
    memset(b,true,sizeof(b));
    memset(p,0,sizeof(p));
    b[0]=b[1]=false;
    for(long long i=2;i<=maxn;i++)
    {
        if(b[i])
        {
            p[k++]=i;
            for(int j=i+i;j<=maxn;j+=i) b[j]=false;
        }
    }
}

int phi(long long tmp)
{
    int l=0,r=k;
    while(l<=r)
    {
        long long mid=(l+r)/2;
        if(p[mid]>tmp) r=mid-1;
        else l=mid+1;
    }
     for(int i=max(r,0);;i++)
        if(p[i]>tmp)
        return p[i];
}

int main()
{
    is_prime();

    long long n,m,l,sum=0;
    cin>>n;
    long long ans=1;
    while(n--)
    {
        cin>>m;
        while(m--)
        {
            cin>>l;
            sum+=phi(l);
        }

        printf("Case %lld: %lld Xukha
",ans,sum);
        ans++;
        sum=0;
    }

    return 0;
}