Light OJ 1307 Bi-shoe and Phi-shoe(欧拉函数)
2015-02-16 17:53
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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.
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].
题意:求欧拉phi值大于等于i的x的和的最小值。
将欧拉phi求出来然后暴力遍历即可。。
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 |
将欧拉phi求出来然后暴力遍历即可。。
#include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include<string> #include<iostream> #include<queue> #include<cmath> #include<map> #include<stack> #include<bitset> using namespace std; #define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i ) #define REP( i , n ) for ( int i = 0 ; i < n ; ++ i ) #define CLEAR( a , x ) memset ( a , x , sizeof a ) typedef long long LL; typedef pair<int,int>pil; const int INF = 0x3f3f3f3f; const int maxn=1e6+5; int phi[maxn+10]; int num[maxn]; int t,n; void init() { REPF(i,2,maxn) phi[i]=0; phi[1]=1; REPF(i,2,maxn) if(!phi[i]) for(int j=i;j<=maxn;j+=i) { if(!phi[j]) phi[j]=j; phi[j]=phi[j]/i*(i-1); } } int main() { init(); scanf("%d",&t); int cas=1; while(t--) { scanf("%d",&n); REP(i,n) scanf("%d",&num[i]); sort(num,num+n); LL sum=0; for(int i=0,j=2;i<n;) { if(phi[j]>=num[i]) { sum+=j; i++; } else j++; } printf("Case %d: %lld Xukha\n",cas++,sum); } return 0; }
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