LightOJ 1370 Bi-shoe and Phi-shoe(素数筛)
Click Here~~ 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 题目大意: 解题思路: #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<<" Xukhan";
}
return 0;
}
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