LightOJ-1370 Bi-shoe and Phi-shoe (欧拉函数+二分)
Problem Description
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.
SampleInput
3
5
1 2 3 4 5
6
10 11 12 13 14 15
2
1 1
SampleOutput
Case 1: 22 Xukha
Case 2: 88 Xukha
Case 3: 4 Xukha
题意:
T组数据,每组数据给定1<=n<=1e4,接下来n个整数1<=xi<=1e6,对于每一个xi,找出最小的yi,使euler(yi) >= xi。求yi之和。
思路:
由欧拉函数定义可知,euler(x) < x。所以对每一个x,找大于它的第一个质数即为答案。线性筛法打素数表,二分查找。
AC代码:92MS
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 #include <cstdlib> 5 #include <algorithm> 6 using namespace std; 7 typedef long long ll; 8 9 const int MAXN = 1e6 + 15; 10 const int PRIME_MAX = MAXN / 2; 11 bool vis[MAXN]; 12 int prime[PRIME_MAX], top; 13 14 void init() { 15 ll i, j; 16 top = 0; 17 vis[1] = false; 18 for(i = 2; i < MAXN; ++i) { 19 if(!vis[i]) { 20 prime[top++] = i; 21 } 22 for(j = 0; prime[j] * i < MAXN; ++j) { 23 vis[prime[j] * i] = true; 24 if(i % prime[j] == 0) 25 break; 26 } 27 } 28 } 29 30 int main() { 31 init(); 32 int t, n, i, now, k; 33 scanf("%d", &t); 34 for(k = 1; k <= t; ++k) { 35 ll ans = 0; 36 scanf("%d", &n); 37 for(i = 1; i <= n; ++i) { 38 scanf("%d", &now); 39 ans += *upper_bound(prime, prime + top, now); 40 } 41 printf("Case %d: %lld Xukha\n", k, ans); 42 } 43 return 0; 44 }