LightOJ1370 Bishoe and Phishoe (欧拉函数+二分)

Problem 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.

SampleInput

3

5

1 2 3 4 5

6

10 11 12 13 14 15

2

1 1

SampleOutput

Case 1: 22 Xukha

Case 2: 88 Xukha

Case 3: 4 Xukha

 

题意：

T组数据，每组数据给定1<=n<=1e4，接下来n个整数1<=xi<=1e6，对于每一个xi，找出最小的yi，使euler(yi) >= xi。求yi之和。

思路：

由欧拉函数定义可知，euler(x) < x。所以对每一个x，找大于它的第一个质数即为答案。线性筛法打素数表，二分查找。

AC代码：92MS

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
using namespace std;
typedef long long ll;

const int MAXN = 1e6 + 15;
const int PRIME_MAX = MAXN / 2;
bool vis[MAXN];
int prime[PRIME_MAX], top;

void init() {
    ll i, j;
    top = 0;
    vis[1] = false;
    for(i = 2; i < MAXN; ++i) {
        if(!vis[i]) {
            prime[top++] = i;
        }
        for(j = 0; prime[j] * i < MAXN; ++j) {
            vis[prime[j] * i] = true;
            if(i % prime[j] == 0)
                break;
        }
    }
}

int main() {
    init();
    int t, n, i, now, k;
    scanf("%d", &t);
    for(k = 1; k <= t; ++k) {
        ll ans = 0;
        scanf("%d", &n);
        for(i = 1; i <= n; ++i) {
            scanf("%d", &now);
            ans += *upper_bound(prime, prime + top, now);
        }
        printf("Case %d: %lld Xukha\n", k, ans);
    }
    return 0;
}