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
题意:
给一些数Ai(第 i 个数),Ai这些数代表的是某个数欧拉函数的值,我们要求出数 Ni 的欧拉函数值不小于Ai。而我们要求的就是这些 Ni 这些数字的和sum,而且我们想要sum最小,求出sum最小多少。
解题思路:
要求和最小,我们可以让每个数都尽量小,那么我们最后得到的肯定就是一个最小值。
给定一个数的欧拉函数值ψ(N),我们怎么样才能求得最小的N?
我们知道,一个素数P的欧拉函数值ψ(P)=P-1。所以如果我们知道ψ(N),那么最小的N就是最接近ψ(N),并且大于ψ(N)的素数。我们把所有素数打表之后再判断就可以了。
渣B的二分查找
#pragma comment(linker,"/STACK:102400000,102400000" #include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #include<iostream> #include<algorithm> #include<vector> #include<map> #include<queue> #include<stack> #include<string> #include<map> #include<set> #include<ctime> #define eps 1e-6 #define MAX 1000010 #define INF 0x3f3f3f3f #define LL long long #define pii pair<int,int> #define rd(x) scanf("%d",&x) #define rd2(x,y) scanf("%d%d",&x,&y) ///map<int,int>mmap; ///map<int,int >::iterator it; using namespace std; bool isprm[MAX]; int prm[100000],cnt=0; void isprime() { memset(isprm,1,sizeof(isprm)); isprm[0]=isprm[1]=false; for(int i=2; i<MAX; ++i) { if(isprm[i]) { for(int j=2*i; j<MAX; j+=i) isprm[j]=false; prm[cnt++]=i; } } } int binsear(int tmp) { int l=0,r=cnt; while(l<=r) { int mid=(l+r)/2; if (prm[mid] > tmp) r = mid - 1; else l=mid + 1; } for(int i=max(r,0);;i++) if(prm[i]>tmp) return prm[i]; } int main () { isprime(); int T,n,Case=1; rd(T); while(T--) { rd(n); LL sum=0,tmp; for(int i=0; i<n; i++) { rd(tmp); sum+=binsear(tmp); //cout<<sum<<' '; } printf("Case %d: %lld Xukha\n",Case++,sum); } return 0; }
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