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
代码:
欧拉筛
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<cmath>
#include<stdlib.h>
const int maxn=1e6+5;
typedef long long ll;
using namespace std;
int prime[1000005];
bool vis[1000005];
void erla() {
int cnt =0;
memset(vis,false,sizeof(vis));
memset(prime,0,sizeof(prime));
for(int t=2; t<=1000005; t++) {
if(!vis[t]) {
prime[cnt++]=t;
}
for(int j=0; j<cnt&&t*prime[j]<=1000005; j++) {
vis[t*prime[j]]=true;
if(t%prime[j]==0) {
break;
}
}
}
}
int main()
{
int T;
cin>>T;
erla();
int n;
int cnt=1;
while(T--)
{
cin>>n;
ll sum=0;
int x;
int l,r;
for(int t=0;t<n;t++)
{
scanf("%d",&x);
for(int j=x+1;j<=1000003;j++)
{
if(vis[j]==false)
{
sum+=j;
break;
}
}
}
cout<<"Case "<<cnt<<": "<<sum<<" Xukha"<<endl;
cnt++;
}
return 0;
}二分+欧拉函数
wa了，没找到错
代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<cmath>
const int maxn=1e6+5;
typedef long long ll;
using namespace std;
struct node
{
ll val,id;
}p[maxn];
int Eular(int n)
{
int ans = n;
for (int i = 2 ; i * i <= n ; i++)
{
if (n % i == 0)
{
ans -= ans / i;
while (n % i == 0)
n /= i;
}
}
if (n > 1)
ans -= ans / n;
return ans;
}
bool cmp(node x,node y)
{
if(x.val!=y.val)
{
return x.val<y.val;
}
else
{
return x.id<y.id;
}
}
int main()
{
int cc=0;
for(int t=2;t<=1000003;t++)
{
p[cc].val=Eular(t);
p[cc++].id=t;
}
sort(p,p+cc,cmp);
int T;
cin>>T;
int n;
int cnt=1;
while(T--)
{
cin>>n;
ll sum=0;
int x;
int l,r;
for(int t=0;t<n;t++)
{
scanf("%d",&x);
l=0;r=cc;
while(l<r)
{
int mid=(l+r)>>1;
if(p[mid].val<x)
{
l=mid+1;
}
else
{
r=mid;
}
}
sum+=p[(l+r)>>1].id;
}
cout<<"Case "<<cnt<<": "<<sum<<" Xukha"<<endl;
cnt++;
}
return 0;
}