| 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 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].
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 |
给一些数Ai（第 i 个数），Ai这些数代表的是某个数欧拉函数的值，我们要求出数 Ni 的欧拉函数值不小于Ai。而我们要求的就是这些 Ni 这些数字的和sum，而且我们想要sum最小，求出sum最小多少。
解题思路：
要求和最小，我们可以让每个数都尽量小，那么我们最后得到的肯定就是一个最小值。
给定一个数的欧拉函数值ψ(N)，我们怎么样才能求得最小的N?
我们知道，一个素数P的欧拉函数值ψ(P)=P-1。所以如果我们知道ψ(N)，那么最小的N就是最接近ψ(N)，并且大于ψ(N)的素数。我们把所有素数打表之后再判断就可以了。
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 Xukha\n",xp++,ans); } return 0; }
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LightOJ Bi-shoe and Phi-shoe 1370【欧拉函数+素数打表】
原文地址：http://blog.csdn.net/ydd97/article/details/47828771
