lightoj 1370欧拉函数
Time Limit: 2 second(s) | Memory Limit: 32 MB |
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 |
Output for Sample Input |
3 5 1 2 3 4 5 6 10 11 12 13 14 15 2 1 1 |
Case 1: 22 Xukha Case 2: 88 Xukha Case 3: 4 Xukha
|
题意:每一个数字的欧拉函数要大于或等于该数字。求,最小的欧拉函数的下标和的大小。
答案要用longlong存。
1 #include<cstdio> 2 using namespace std; 3 const int maxn=1000006; 4 int prime[maxn]; 5 6 void init() 7 { 8 for(int i=1;i<maxn;i++) 9 prime[i]=i; 10 for(int i=2;i<maxn;i++){ 11 if(prime[i]==i){ 12 int temp=i; 13 for(int j=i;j<maxn;j+=temp){ 14 prime[j]=prime[j]/temp*(temp-1); 15 } 16 } 17 } 18 } 19 int main() 20 { 21 int n,T; 22 init(); 23 scanf("%d",&T); 24 for(int kkk=1;kkk<=T;kkk++){ 25 scanf("%d",&n); 26 long long ans=0; 27 while(n--){ 28 int x,temp; 29 scanf("%d",&x); 30 temp=x+1; 31 while(prime[temp]<x) 32 temp++; 33 ans+=temp; 34 } 35 printf("Case %d: %lld Xukha\n",kkk,ans); 36 } 37 38 return 0; 39 }