1370 - Bi-shoe and Phi-shoe

1370 - Bi-shoe and Phi-shoe

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
3 5 1 2 3 4 5 6 10 11 12 13 14 15 2 1 1	Case 1: 22 Xukha Case 2: 88 Xukha Case 3: 4 Xukha

 

 

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思路：素数打表，欧拉函数，二分。

由于要选最小的，所以假如ola[y]<ola[x]（x<y）那么我们要选的是x，所以如果后面的小于前面的，我们直接把后面的更新为前面的这样欧拉函数值才会是呈递增，那没用二分选取就行了。

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<map>
#include<set>
using namespace std;
bool prime[3300000+5];
int su[300000];
typedef long long LL;
typedef struct pp
{
        int x;
        int id;
} ss;
ss ola[3300000+5];
int aa[10005];
bool cmp(struct pp nn,struct pp mm)
{
        if(nn.x==mm.x)
                return nn.id<mm.id;
        else return nn.x<mm.x;
}
typedef unsigned long long  ll;
int main(void)
{
        int i,j,k;
        for(i=2; i<=7000; i++)
        {
                if(!prime[i])
                        for(j=i; i*j<=(3300000); j++)
                        {
                                prime[i*j]=true;
                        }
        }
        int ans=0;
        for(i=2; i<=(3300000); i++)
                if(!prime[i])
                        su[ans++]=i;
        for(i=1; i<=3300000; i++)
        {
                ola[i].id=i;
                ola[i].x=i;
        }
        for(i=0; i<ans; i++)
        {
                for(j=1; su[i]*j<=3300000; j++)
                {
                        ola[su[i]*j].x=ola[su[i]*j].x/(su[i])*(su[i]-1);
                }
        }
        for(i=2; i<3300000; i++)
        {
                if(ola[i].x>ola[i+1].x)
                {
                        ola[i+1].x=ola[i].x;
                }
        }
        scanf("%d",&k);
        int s;
        int p,q;
        for(s=1; s<=k; s++)
        {
                LL sum=0;
                scanf("%d",&p);
                for(i=0; i<p; i++)
                {
                        scanf("%d",&aa[i]);
                        int l=2;
                        int r=4*1000000;
                        int zz=0;
                        while(l<=r)
                        {
                                int mid=(l+r)>>1;
                                if(ola[mid].x<aa[i])
                                {
                                        l=mid+1;
                                }
                                else
                                {
                                        zz=mid;
                                        r=mid-1;
                                }
                        }
                        sum+=ola[zz].id;
                }
                printf("Case %d: %lld Xukha\n",s,sum);
        }
        return 0;
}