lightoj 1370 - Bi-shoe and Phi-shoe（欧拉函数）

最新推荐文章于 2020-09-04 17:13:24 发布

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本篇介绍了一个有趣的竹竿跳高问题，通过使用欧拉函数来筛选出满足条件的竹竿长度，实现成本最小化的目标。文章详细解释了如何通过预处理和二分搜索来高效解决此问题。

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

1 2 3 4 5

10 11 12 13 14 15

1 1

Case 1: 22 Xukha

Case 2: 88 Xukha

Case 3: 4 Xukha

题意就是给出n个整数，每个整数ai找出最小的xi使得Φ(xi)>=ai,求x1+x2+...+xn。

Φ(xi)就是1~（xi-1）中有多少个数与xi互质。

利用欧拉函数筛先把每个数的欧拉函数值筛出来（注意maxn一定要设大一点，要保证maxn欧拉函数的值比ai的最大值要大），然后再预处理一波花费xi最多能得到多少价值，这样花费和价值就是单调函数了，就可以利用二分求出最小的花费。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long

using namespace std;

const int maxn = 3e6;
int phi[maxn+10];
void phi_table()
{
    int i,j;
    phi[1]=0;
    for(i=2;i<=maxn;i++)
    {
        if(phi[i])
            continue;
        for(j=i;j<=maxn;j+=i)
        {
            if(!phi[j])
                phi[j]=j;
            phi[j]=phi[j]/i*(i-1);
        }
    }
    for(i=2;i<=maxn;i++)
    {
        phi[i] = max(phi[i],phi[i-1]);
    }
}

int binarysearch(int x)
{
    int l = 1,r = maxn;
    int ans = r;
    while(l <= r)
    {
        int mid = (l+r)/2;
        if(phi[mid] >= x)
        {
            r = mid - 1;
            ans = mid;
        }
        else
            l = mid + 1;
    }

    return ans;
}
int main(void)
{
    int T,n,i,j;
    phi_table();
    scanf("%d",&T);
    int cas = 1;
    while(T--)
    {
        scanf("%d",&n);
        LL sum = 0;
        for(i=1;i<=n;i++)
        {
            int x;
            scanf("%d",&x);
            sum += binarysearch(x);
        }
        printf("Case %d: %lld Xukha\n",cas++,sum);
    }

    return 0;
}