题目链接：https://vjudge.net/problem/LightOJ-1370
| 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
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：
给出n个数，为每个数x找到满足：x<=Euler(y) 的最小的y，其中Euler()为欧拉函数。
题解：
可知，当y为素数，Euler(y) = y-1。所以，只需从x+1开始，找到第一个素数即可。
代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+;
const int MAXN = 1e6+; bool vis[MAXN]; void getPrime()
{
int m = (int)sqrt(MAXN);
memset(vis, , sizeof(vis));
vis[] = ;
for(int i = ; i<=m; i++) if(!vis[i])
for(int j = i*i; j<MAXN; j+=i)
vis[j] = ;
} int main()
{
getPrime();
int T, n;
scanf("%d", &T);
for(int kase = ; kase<=T; kase++)
{
LL ans = ;
scanf("%d", &n);
for(int i = ; i<=n; i++)
{
int x;
scanf("%d", &x);
for(int j = x+;;j++) if(!vis[j]){
ans += j;
break;
}
}
printf("Case %d: %lld Xukha\n", kase,ans);
}
}