Ligh OJ 1370 Party All the Time (欧拉函数 +素数打表)
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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.
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:
给你一些数,把每一个数,看成一个数的欧拉函数值。如一个欧拉函数值为x,那么他对应这初始值为y因为很多数有相同的欧拉函数值,即一个x对应着很多个y,我们求出数Y欧拉函数不小于x。问:这些y值和最小为多少。
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。
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int maxn=1000000+1000;int is_prime[maxn];int n;void init(){ memset(is_prime,0,sizeof(is_prime)); is_prime[1]=1; for(LL i=2;i<maxn;i++) { if(!is_prime[i]) { for(LL j=i*i;j<maxn;j+=i) is_prime[j]=1; } }}int main(){ init(); int t,x; int cas=0; scanf("%d",&t); while(t--) { long long ans=0; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&x); for(int j=x+1;;j++) { if(!is_prime[j]){ ans+=j; break; } } } printf("Case %d: %lld Xukha\n",++cas,ans); } return 0;}
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