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A - Bi-shoe and Phi-shoe

Time Limit: 2000 MS Memory Limit: 32768 KB 64bit IO Format: %lld &
%llu

Submit Status

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 6 ] .

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

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <ctype.h>
#include <time.h>
#include <queue>
#include <iterator>

using namespace std;

const int MAXN = 1000100;
int n, m, t;
int prime[MAXN], ph[MAXN], p[MAXN], q[MAXN];

void solve(int n)
{
    memset(prime,0,sizeof(prime));
    memset(p,0,sizeof(p));
    memset(ph,0,sizeof(ph));
    int t = 0;

    for (int i = 2; i <= n; i++)
    {
        if (p[i] == 0)
            prime[++t] = i;
        for (int j = i * 2; j <= n; j+=i)
        {
            p[j] = 1;
        }
    }
    t = 1;
    for (int i = 1; i <= n; i++)
    {
        while (i >= prime[t])
        {
            t++;
        }
        if (i < prime[t])
            ph[i] = prime[t];
    }
}
/*
int phi[MAXN];

void Phi(int n)
{
    for (int i = 0; i <= n; i++)
        phi[i] = 0;
    phi[1] = 1;
    for (int i = 2; i <= n; i++)
    {
        if (!phi[i])
        {
            for (int j = i; j <= n; j += i)
            {
                if (!phi[j]) phi[j] = j;
                phi[j] = phi[j] / i*(i-1);
            }
        }
    }
}
*/
int main()
{
    int cases = 1;
    solve(1001000);
    scanf("%d",&t);
    while (t--)
    {
        long long ans = 0;
        scanf("%d",&n);
        for (int i = 0; i < n; i++)
            scanf("%d", &q[i]);

        sort(q,q+n);

        for (int i = 0; i < n; i++)
            ans += ph[q[i]];
        printf("Case %d: %lld Xukha\n",cases++,ans);
    }
    return 0;
}