Lightoj 1370 素数打表 +二分

1370 - Bi-shoe and Phi-shoe

PDF (English)

Statistics

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：给你一个数组a，找到一个最小的值x，使得phi（x）>=phi（a[i]）；求x的和最小

思路：根据欧拉函数，一个素数p的欧拉函数值为p-1；所以最小的数为大于这个数的最小素数；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define esp 0.00000000001
const int N=1e3+10,M=1e6+1000,inf=1e9+10,mod=1000000007;
const int MAXN=1000010;
int prime[MAXN];//保存素数
bool vis[MAXN];//初始化
int Prime(int n)
{
    int cnt=0;
    memset(vis,0,sizeof(vis));
    for(int i=2;i<n;i++)
    {
        if(!vis[i])
        prime[cnt++]=i;
        for(int j=0;j<cnt&&i*prime[j]<n;j++)
        {
            vis[i*prime[j]]=1;
            if(i%prime[j]==0)
            break;
        }
    }
    return cnt;
}
int main()
{
    int cnt=Prime(MAXN);
    int x,y,z,i,t;
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&x);
        ll ans=0;
        for(i=0;i<x;i++)
        {
            scanf("%d",&y);
            int st=0;
            int en=cnt-1;
            while(st<en)
            {
                int mid=(st+en)>>1;
                if(prime[mid]<=y)
                st=mid+1;
                else
                en=mid;
            }
            ans+=prime[st];
        }
        printf("Case %d: %lld Xukha\n",cas++,ans);
    }
    return 0;
}