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
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
using namespace std;
typedef long long ll; const int MAXN = 1e6 + ;
const int PRIME_MAX = MAXN / ;
bool vis[MAXN];
int prime[PRIME_MAX], top; void init() {
ll i, j;
top = ;
vis[] = false;
for(i = ; i < MAXN; ++i) {
if(!vis[i]) {
prime[top++] = i;
}
for(j = ; prime[j] * i < MAXN; ++j) {
vis[prime[j] * i] = true;
if(i % prime[j] == )
break;
}
}
} int main() {
init();
int t, n, i, now, k;
scanf("%d", &t);
for(k = ; k <= t; ++k) {
ll ans = ;
scanf("%d", &n);
for(i = ; i <= n; ++i) {
scanf("%d", &now);
ans += *upper_bound(prime, prime + top, now);
}
printf("Case %d: %lld Xukha\n", k, ans);
}
return ;
}
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