LightOJ Bi-shoe and Phi-shoe 1370【欧拉函数+素数打表】

1370 - Bi-shoe and Phi-shoe

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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	Output for Sample Input
3 5 1 2 3 4 5 6 10 11 12 13 14 15 2 1 1	Case 1: 22 Xukha Case 2: 88 Xukha Case 3: 4 Xukha

 

题意：

给一些数Ai（第 i 个数），Ai这些数代表的是某个数欧拉函数的值，我们要求出数 Ni 的欧拉函数值不小于Ai。而我们要求的就是这些 Ni 这些数字的和sum，而且我们想要sum最小，求出sum最小多少。

解题思路：

要求和最小，我们可以让每个数都尽量小，那么我们最后得到的肯定就是一个最小值。

给定一个数的欧拉函数值ψ(N)，我们怎么样才能求得最小的N?

我们知道，一个素数P的欧拉函数值ψ(P)=P-1。所以如果我们知道ψ(N)，那么最小的N就是最接近ψ(N)，并且大于ψ(N)的素数。我们把所有素数打表之后再判断就可以了。

AC代码：

#include <stdio.h>
#include <math.h>
#include <vector>
#include <queue>
#include <string>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int MAXN = 1000000+100;

LL is_prime[MAXN];
LL prime[MAXN];

void find_prime()
{
    is_prime[1]=1;
    for(int i=2;i<MAXN;i++){
        if(!is_prime[i]){
            for(int j=i*2;j<MAXN;j+=i)
                is_prime[j]=1;
        }
    }
}

int main()
{
    find_prime();
    int t;
    scanf("%d",&t);
    int xp=1;
    while(t--){
        int n;
        scanf("%d",&n);
        int num;
        LL ans=0;
        for(int i=0;i<n;i++){
            scanf("%d",&num);
            for(int j=num+1;;j++)
                if(!is_prime[j]){
                    ans+=j;
                    break;
                }
        }
        printf("Case %d: %lld Xukha\n",xp++,ans);
    }
    return 0;
}