LightOJ 1370 Bi-shoe and Phi-shoe【欧拉打表】

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：
n个数，满足 phi(x)>=y $phi(x)>=y$,（x为满足条件的最小值），将x值求和。
分析：
euler打表，1e6~2e6之间一定有个质数，就会满足条件，打表时把MAXN数据开大点就行。

#include <cstdio>
#include <cmath>
typedef long long LL;

const LL MAXN = 1e6 + 1e4;
LL eul[MAXN],p[MAXN];

void init() {
    for(LL i = 2; i < MAXN; i++) { //范围放大点，总会遇到一个质数
        if(eul[i]) continue;
        for(LL j = i; j < MAXN; j += i) {
            if(!eul[j]) eul[j] = j;
            eul[j] = eul[j] * (i - 1) / i;
        }
    }
    for(LL i = 1; i <= 1e6; i++) {
        for(LL j = i; ; j++) {
            if(eul[j] >= i) {
                p[i] = j;
                break;
            }
        }
    }
}

int main() {
    init();
    LL T,n,t = 1;
    scanf("%lld",&T);
    while(T--) {
        LL sum = 0,k;
        scanf("%lld",&n);
        for(LL i = 1; i <= n; i++) {
            scanf("%lld",&k);
            sum += p[k];
        }
        printf("Case %lld: %lld Xukhan",t++,sum);
    }
    return 0;
}