lightoj 1370

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 longusing 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;}