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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, 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
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;
}
BNU 13288 Bi-shoe and Phi-shoe 【素数筛选】
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原文地址：http://blog.csdn.net/u014427196/article/details/44156609
