1370 - Bi-shoe and Phi-shoe

阅读原文时间：2023年07月09日阅读：1

1370 - Bi-shoe and Phi-shoe

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

1 2 3 4 5

10 11 12 13 14 15

1 1

Case 1: 22 Xukha

Case 2: 88 Xukha

Case 3: 4 Xukha

思路：素数打表，欧拉函数，二分。

由于要选最小的，所以假如ola[y]<ola[x]（x<y）那么我们要选的是x，所以如果后面的小于前面的，我们直接把后面的更新为前面的这样欧拉函数值才会是呈递增，那没用二分选取就行了。

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
bool prime[3300000+5];
int su[300000];
typedef long long LL;
typedef struct pp
{
int x;
int id;
} ss;
ss ola[3300000+5];
int aa[10005];
bool cmp(struct pp nn,struct pp mm)
{
if(nn.x==mm.x)
return nn.idola[i+1].x)
{
ola[i+1].x=ola[i].x;
}
}
scanf("%d",&k);
int s;
int p,q;
for(s=1; s<=k; s++) { LL sum=0; scanf("%d",&p); for(i=0; i>1;
if(ola[mid].x<aa[i])
{
l=mid+1;
}
else
{
zz=mid;
r=mid-1;
}
}
sum+=ola[zz].id;
}
printf("Case %d: %lld Xukha\n",s,sum);
}
return 0;
}

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