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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
题意 ： 给你 n 个数，分别找出大于等于这些数的欧拉函数值的 x ,使 x 的和最小
//我的思路：首先欧拉函数打表，再分别找出最小值
//之前纠结于 1 1 得出 4 的这组数据 后来看见 Φ (n) = numbers less than n which are relatively prime
//这个题中的规定和欧拉函数不一样 欧拉函数是小于等于 n , 所以在本题中 1 的值 不为 1 而是 0；
#include <iostream> #include <algorithm> #include <cstdio> #define ll long long using namespace std; #define max 1123456 ll eular[max]; void init() { eular[1]=1; for(int i=2;i<max;i++) eular[i]=i; for(int i=2;i<max;i++) { if(eular[i]==i) { for(int j=i;j<max;j+=i) eular[j]=eular[j]/i*(i-1); } } } int main() { ll c,data[10005],ans,tmp=0; cin>>c; init(); while(c--) { int n; tmp++; cin>>n; ans=0; for(int i=0;i<n;i++) cin>>data[i]; for(int i=0;i<n;i++) { for(int j=data[i]+1;j<max;j++) if(eular[j]>=data[i]) { ans+=j; break; } } printf("Case %lld: %lld Xukha\n",tmp,ans); } return 0; }
A - Bi-shoe and Phi-shoe (欧拉函数打表)
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原文地址：http://www.cnblogs.com/nefu929831238/p/5734287.html
