LightOJ Bi-shoe and Phi-shoe 1370【欧拉函数+素数打表】
1370 - Bi-shoe and Phi-shoe
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. InputInput 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]. OutputFor each case,print the case number and the minimum possible money spent for buying the bamboos. See the samples for details.
? 题意:
解题思路:
AC代码: #include <stdio.h> #include <math.h> #include <vector> #include <queue> #include <string> #include <string.h> #include <stdlib.h> #include <iostream> #include <algorithm> using namespace std; typedef long long LL; const int MAXN = 1000000+100; LL is_prime[MAXN]; LL prime[MAXN]; void find_prime() { is_prime[1]=1; for(int i=2;i<MAXN;i++){ if(!is_prime[i]){ for(int j=i*2;j<MAXN;j+=i) is_prime[j]=1; } } } int main() { find_prime(); int t; scanf("%d",&t); int xp=1; while(t--){ int n; scanf("%d",&n); int num; LL ans=0; for(int i=0;i<n;i++){ scanf("%d",&num); for(int j=num+1;;j++) if(!is_prime[j]){ ans+=j; break; } } printf("Case %d: %lld Xukhan",xp++,ans); } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |