Crawling in process... Crawling failed 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, 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
题目大意：这道题本质上的意思就是给你一个数N,让你寻找最小的k满足&(k)>=N(&指的是欧拉函数）
思路分析：考察了欧拉函数的简单性质，即满足&(k)>=N的最小数为N+1Z之后的第一个素数
代码：
#include<iostream>
#include<cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long ll;
const int maxn=1e6+;
int phi[maxn];
int prime[maxn];
bool check[maxn];
int tot;
void make_phi()
{
tot=;
memset(check,true,sizeof(check));
phi[]=;
for(int i=;i<=maxn;i++)
{
if(check[i])
{
prime[tot++]=i;
phi[i]=i-;
}
for(int j=;j<tot&&i*prime[j]<=maxn;j++)
{
check[i*prime[j]]=false;
if(i%prime[j]==)
{
phi[i*prime[j]]=phi[i]*prime[j];
break;
}
else prime[i*prime[j]]=phi[i]*(prime[j]-);
}
}
}
int kase;
int main()
{
int T;
make_phi();
scanf("%d",&T);
kase=;
ll num;
while(T--)
{
int n;
scanf("%d",&n);
ll ans=;
while(n--)
{
scanf("%lld",&num);
ll k=num+;
for(ll i=k;;i++)
{
if(check[i])
{
ans+=i;
break;
}
}
}
printf("Case %d: %lld Xukha\n",++kase,ans);
}
}