1370 - Bi-shoe and Phi-shoe

PDF (English)

Statistics

Forum

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.

题意：

给你一些数，把每一个数，看成一个数的欧拉函数值。如一个欧拉函数值为x，那么他对应这初始值为y因为很多数有相同的欧拉函数值，即一个x对应着很多个y，我们求出数Y欧拉函数不小于x。问：这些y值和最小为多少。

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。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn=1000000+1000;
int is_prime[maxn];
int n;
void init()
{
    memset(is_prime,0,sizeof(is_prime));
    is_prime[1]=1;
    for(LL i=2;i<maxn;i++)
    {
        if(!is_prime[i])
        {
            for(LL j=i*i;j<maxn;j+=i)
                is_prime[j]=1;
        }
    }
}
int main()
{
    init();
    int t,x;
    int cas=0;
    scanf("%d",&t);
    while(t--)
    {
        long long ans=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&x);
            for(int j=x+1;;j++)
            {
                if(!is_prime[j]){
                    ans+=j;
                    break;
                }
            }
        }
        printf("Case %d: %lld Xukha\n",++cas,ans);
    }
    return 0;
}

版权声明：本文为博主原创文章，转载记得著名出处，谢谢！