light_oj 1370 欧拉函数+二分

A - Bi-shoe and Phi-shoe

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

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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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<string>
#include<math.h>
#include<cctype>

using namespace std;

typedef long long ll;
const int maxn=2000100;
const int INF=(1<<29);
const double EPS=0.0000000001;
const double Pi=acos(-1.0);

int n;
int euler[maxn];

void Init()
{
    euler[1]=1;
    for(int i=2;i<maxn;i++) euler[i]=i;
    for(int i=2;i<maxn;i++){
        if(euler[i]==i){
            for(int j=i;j<maxn;j+=i) euler[j]=euler[j]/i*(i-1);
        }
    }
}

int main()
{
    int T;cin>>T;
    int tag=1;
    Init();
    for(int i=2;i<maxn;i++) euler[i]=max(euler[i],euler[i-1]);
    while(T--){
        cin>>n;
        ll ans=0;
        while(n--){
            int t;
            scanf("%d",&t);
            ans+=lower_bound(euler+2,euler+maxn,t)-euler;
        }
        printf("Case %d: %lld Xukha\n",tag++,ans);
    }
    return 0;
}