LightOJ1370 Bi-shoe and Phi-shoe 欧拉函数筛法

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

欧拉筛，输出需用long long。

#include <iostream>#include <cstdio>#include <map>#include <cmath>#include <map>#include <vector>#include <algorithm>#include <cstring>#include <string>using namespace std;#define LL long long#define maxn 1000009LL euler[maxn];void euler_init(){    int i,j;    euler[1]=1;    for(i=2;i<maxn;i++)      euler[i]=i;    for(i=2;i<maxn;i++)       if(euler[i]==i)          for(j=i;j<maxn;j+=i)            euler[j]=euler[j]/i*(i-1);}int main(){    int t,n,ca=1,a;    euler_init();    scanf("%d",&t);    while(t--){        LL ans=0;        scanf("%d",&n);        for(int i=0;i<n;i++){            scanf("%d",&a);            int j=a+1;            while(euler[j]<a){                j++;            }            ans+=j;        }        printf("Case %d: %lld Xukha\n",ca++,ans);    }    return 0;}