| Time Limit: 2000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
Description
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, 10].
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
3
5
1 2 3 4 5
6
10 11 12 13 14 15
2
1 1
Sample Output
Case 1: 22 Xukha
Case 2: 88 Xukha
Case 3: 4 Xukha
Source
/**
题意：f(n) 表示 小于等于 n 的数中素数的个数；给出一串数 比如x 求f(x) >= x 的最小和
做法：欧拉函数
**/
#include <iostream>
#include<cmath>
#include<string.h>
#include<stdio.h>
#include<algorithm>
#include<stack>
#define maxn 1000000 + 10
int mindiv[maxn],phi[maxn],sum[maxn];
int mmap[maxn];
int mmpp[maxn];
using namespace std;
void solve()
{
for(int i=; i<maxn; i++)
{
mindiv[i] = i;
}
for(int i=; i*i<maxn; i++)
{
if(mindiv[i] == i)
{
for(int j=i*i; j<maxn; j+=i)
{
mindiv[j] = i;
}
}
}
phi[] = ;
for(int i=; i<maxn; i++)
{
phi[i] = phi[i/mindiv[i]];
if((i/mindiv[i])%mindiv[i] == )
{
phi[i] *=mindiv[i];
}
else
{
phi[i] *= mindiv[i] -;
}
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
int T;
scanf("%d",&T);
int n;
solve();
int Case = ;
while(T--)
{
scanf("%d",&n);
int res = ;
long long sum = ;
for(int i=; i<n; i++)
{
scanf("%d",&mmpp[i]);
}
sort(mmpp,mmpp+n);
int tt= ,Index = ;
int j = ;
phi[] = ;
for(int i=; i<n; )
{
if(phi[j] >= mmpp[i])
{
sum += j;
i++;
}
else j++;
}
printf("Case %d: %lld Xukha\n",Case++,sum);
}
return ;
}