Bi-shoe and Phi-shoe（欧拉函数变形）

Bi-shoe and Phi-shoe
Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu
Submit Status Practice LightOJ
1370

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

题意：

给一些数Ai（第 i 个数），Ai这些数代表的是某个数欧拉函数的值，我们要求出数 Ni 的欧拉函数值不小于Ai。而我们要求的就是这些 Ni 这些数字的和sum，而且我们想要sum最小，求出sum最小多少。

解题思路：

要求和最小，我们可以让每个数都尽量小，那么我们最后得到的肯定就是一个最小值。

给定一个数的欧拉函数值ψ(N)，我们怎么样才能求得最小的N?

我们知道，一个素数P的欧拉函数值ψ(P)=P-1。所以如果我们知道ψ(N)，那么最小的N就是最接近ψ(N)，并且大于ψ(N)的素数。我们把所有素数打表之后再判断就可以了。

渣B的二分查找

#pragma comment(linker, "/STACK:102400000,102400000"
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<ctime>
#define eps 1e-6
#define MAX 1000010
#define INF 0x3f3f3f3f
#define LL long long
#define pii pair<int,int>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
///map<int,int>mmap;
///map<int,int >::iterator it;
using namespace std;

bool isprm[MAX];
int prm[100000],cnt=0;
void isprime()
{
memset(isprm,1,sizeof(isprm));
isprm[0]=isprm[1]=false;
for(int i=2; i<MAX; ++i)
{
if(isprm[i])
{
for(int j=2*i; j<MAX; j+=i)
isprm[j]=false;
prm[cnt++]=i;
}
}
}
int binsear(int tmp)
{
int l=0,r=cnt;
while(l<=r)
{
int mid=(l+r)/2;
if (prm[mid] > tmp)
r = mid - 1;
else
l=mid + 1;
}
for(int i=max(r,0);;i++)
if(prm[i]>tmp)
return prm[i];
}
int main ()
{
isprime();
int T,n,Case=1;
rd(T);
while(T--)
{
rd(n);
LL sum=0,tmp;
for(int i=0; i<n; i++)
{
rd(tmp);
sum+=binsear(tmp);
//cout<<sum<<' ';
}
printf("Case %d: %lld Xukha\n",Case++,sum);
}
return 0;
}