Bi-shoe and Phi-shoe（欧拉函数变形）

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：

要求和最小，我们可以让每个数都尽量小，那么我们最后得到的肯定就是一个最小值。

给定一个数的欧拉函数值ψ(N)，我们怎么样才能求得最小的N?

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ψ(N)，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。

渣B的二分查找

#pragma comment(linker,"/STACK:102400000,102400000"
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define MAX 1000010
#define INF 0x3f3f3f3f
#define LL long long
#define pii pair<int,int>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
///map<int,int>mmap;
///map<int,int >::iterator it;
using namespace std;

bool isprm[MAX];
int prm[100000],cnt=0;
void isprime()
{
    memset(isprm,1,sizeof(isprm));
    isprm[0]=isprm[1]=false;
    for(int i=2; i<MAX; ++i)
    {
        if(isprm[i])
        {
            for(int j=2*i; j<MAX; j+=i)
                isprm[j]=false;
            prm[cnt++]=i;
        }
    }
}
int binsear(int tmp)
{
    int l=0,r=cnt;
    while(l<=r)
    {
        int mid=(l+r)/2;
        if (prm[mid] > tmp)
            r = mid - 1;
        else
            l=mid + 1;
    }
    for(int i=max(r,0);;i++)
        if(prm[i]>tmp)
        return prm[i];
}
int main ()
{
    isprime();
    int T,n,Case=1;
    rd(T);
    while(T--)
    {
        rd(n);
        LL sum=0,tmp;
        for(int i=0; i<n; i++)
        {
            rd(tmp);
            sum+=binsear(tmp);
            //cout<<sum<<' ';
        }
        printf("Case %d: %lld Xukhan",Case++,sum);
    }
    return 0;
}

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