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A - Bi-shoe and Phi-shoe
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
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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
题目大意：
首先给出T组数据，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 pi 的欧拉函数值不小于ai。而我们要求的就是这些 pi 这些数字的和sum，而且我们想要sum最小，求出sum至少是多少。
解题思路：
要想找到最小的和 只要我们考虑的是每个pi最小 因为素数的欧拉函数值是素数-1，所以我们只需要找到比ai大的第一个素数就行啦，然后我们就想到素数筛啦 哈哈~~ 其实数据范围是1e6，并不是很大 所以可以直接存下
详见代码：
#include <iostream>
#include <cstring>
using namespace std;
typedef long long LL;
const int MAXN = 1e6+5;
int prime[MAXN];///存素数的
int cnt;///素数的个数
bool p[MAXN];///判断素数滴
void isprime()///素数筛
{
cnt = 0;
memset(p, true, sizeof(p));
p[1] = false;
for(LL i=2; i<MAXN; i++)
{
if(p[i])
{
prime[cnt++] = i;
for(LL j=i*i; j<MAXN; j+=i)
p[j] = false;
}
}
}
///没有用的欧拉函数 白写了~~
///int Eular(int m)
///{
/// int ret = m;
/// for(int i=2; i<m; i++)
/// {
/// if(m%i == 0)
/// {
/// ret -= ret/i;
/// while(m%i)
/// m /= i;
/// }
/// }
/// if(m > 1)
/// ret -= ret/m;
/// return ret;
///}
int main()
{
isprime();
///cout<<prime[cnt-1]<<endl;
LL T, m, num;
cin>>T;
for(int cas=1; cas<=T; cas++)
{
cin>>m;
LL sum = 0;
while(m--)
{
cin>>num;
for(int i=num+1; ; i++)///找到第一个素数
{
if(p[i])
{
sum += i;
break;
}
}
}
cout<<"Case "<<cas<<": "<<sum<<" Xukha\n";
}
return 0;
}