LightOJ

1370 - Bi-shoe and Phi-shoe

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

3
5
1 2 3 4 5
6
10 11 12 13 14 15
2
1 1

Output for Sample Input

Case 1: 22 Xukha
Case 2: 88 Xukha
Case 3: 4 Xukha

题意： 给你n个数，问你这每个数的i 满足，eul[x] >= i，的最小的x，然后求和

分析： n的范围是1e5 我们可以将范围内的欧拉值打表后，将每个数i排个序，然后在O(n)的时间内计算出最小值

参考代码

#include<bits/stdc++.h>using namespace std;#define ll long longconst int N = 1e6 + 100;int eul[N<<1];int a[10010];int n;void init() {    for(int i = 1;i < N;i++) {        eul[i] = i;    }    for(int i = 2;i <= N;i++) {        if(eul[i] == i) {            for(int j = i;j <= N;j += i) {                eul[j] = eul[j]/i*(i-1);            }        }    }}ll get() {    ll res = 0;    int r = 0;    for(int i = 2;i < N;i++) {        if(r >= n) break;        if(eul[i] >= a[r]) {            while (eul[i] >= a[r] && r < n) {                res += i*1ll;                r++;            }        }    }    return res;}int main(){    init();    int T;cin>>T;    for(int t = 1;t <= T;t++) {        cin>>n;        for(int i = 0;i < n;i++) cin>>a[i];        sort(a,a+n);        printf("Case %d: %lld Xukha\n",t,get());    }    return 0;}

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