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.
Sample Input
3
5
1 2 3 4 5
6
10 11 12 13 14 15
2
1 1
Sample Output
Case 1: 22 Xukha
Case 2: 88 Xukha
Case 3: 4 Xukha
题目链接：http://www.bnuoj.com/bnuoj/problem_show.php?pid=13288
被这个题目最后的输出格式气得半天，__int64输出用lld输出才AC，用I64d不行。
欧拉函数定义：比他小的并与他互质的个数之和。
题意给你n个luky数字是欧拉函数的值。也就是个数，求和最小，我就想应该是比这个数大的质数就肯定是最小的，这样先把所有质数枚举出来就好办了。用到了素数筛选法。
#include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <iostream> #include <cmath> #include <queue> #include <map> #include <stack> #include <list> #include <vector> #include <ctime> #define LL __int64 #define eps 1e-8 #define pi acos(-1) using namespace std; int a[1010010]; int p[100000]; void INIT(){ memset(a,sizeof(a)); memset(p,sizeof(p)); int t=0; for (int i=2;i<=1010000;i++){ if (a[i]==0) p[++t]=i; for (int j=1,k;j<=t && ((k=i*p[j])<=1010000);j++) { a[k]=1; if (i % p[j]==0) break; } } t=1; for (int i=1;i<=1000001;i++){ if (i<p[t]) a[i]=p[t]; else { t++; a[i]=p[t]; } } } int main(){ INIT(); int T; //freopen("in.txt","r",stdin); //fropen("out.txt","w",stdout) scanf("%d",&T); for (int cas=1;cas<=T;cas++){ int n,k; scanf("%d",&n); LL ans=0; for(int i=1;i<=n;i++){ scanf("%d",&k); ans+=a[k]; } printf("Case %d: %lld Xukha\n",cas,ans); } return 0; }
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