light_oj 1370 欧拉函数+二分
light_oj 1370 欧拉函数+二分
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
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<set> #include<map> #include<string> #include<math.h> #include<cctype> using namespace std; typedef long long ll; const int maxn=2000100; const int INF=(1<<29); const double EPS=0.0000000001; const double Pi=acos(-1.0); int n; int euler[maxn]; void Init() { euler[1]=1; for(int i=2;i<maxn;i++) euler[i]=i; for(int i=2;i<maxn;i++){ if(euler[i]==i){ for(int j=i;j<maxn;j+=i) euler[j]=euler[j]/i*(i-1); } } } int main() { int T;cin>>T; int tag=1; Init(); for(int i=2;i<maxn;i++) euler[i]=max(euler[i],euler[i-1]); while(T--){ cin>>n; ll ans=0; while(n--){ int t; scanf("%d",&t); ans+=lower_bound(euler+2,euler+maxn,t)-euler; } printf("Case %d: %lld Xukha\n",tag++,ans); } return 0; }