Lightoj 1370 素数打表 +二分
2016-08-08 12:15
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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.
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:给你一个数组a,找到一个最小的值x,使得phi(x)>=phi(a[i]);求x的和最小
思路:根据欧拉函数,一个素数p的欧拉函数值为p-1;所以最小的数为大于这个数的最小素数;
PDF (English) | Statistics |
Time Limit: 2 second(s) | Memory Limit: 32 MB |
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 |
思路:根据欧拉函数,一个素数p的欧拉函数值为p-1;所以最小的数为大于这个数的最小素数;
#include<bits/stdc++.h> using namespace std; #define ll long long #define esp 0.00000000001 const int N=1e3+10,M=1e6+1000,inf=1e9+10,mod=1000000007; const int MAXN=1000010; int prime[MAXN];//保存素数 bool vis[MAXN];//初始化 int Prime(int n) { int cnt=0; memset(vis,0,sizeof(vis)); for(int i=2;i<n;i++) { if(!vis[i]) prime[cnt++]=i; for(int j=0;j<cnt&&i*prime[j]<n;j++) { vis[i*prime[j]]=1; if(i%prime[j]==0) break; } } return cnt; } int main() { int cnt=Prime(MAXN); int x,y,z,i,t; int T,cas=1; scanf("%d",&T); while(T--) { scanf("%d",&x); ll ans=0; for(i=0;i<x;i++) { scanf("%d",&y); int st=0; int en=cnt-1; while(st<en) { int mid=(st+en)>>1; if(prime[mid]<=y) st=mid+1; else en=mid; } ans+=prime[st]; } printf("Case %d: %lld Xukha\n",cas++,ans); } return 0; }
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