1370-bi-shoe and Phi-shoe

1370-bi-shoe and Phi-shoe

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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 coaches 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!) is available in the market. According to Xzhila tradition,

Score of a bamboo = Φ (bamboo ' s length)

(Xzhilans is really fond of number theory). For your information, Φ (n) = numbers less than n which is relatively prime (having no common divisor o Ther than 1) to N. So, score of a bamboo of length 9 is 6 as 1, 2, 4, 5, 7, 8 were relatively prime to 9.

The assistant Bi-shoe have to buy one bamboo for each student. As a twist, each pole-vault student of Phi-shoe have a lucky number. Bi-shoe wants to buy bamboos such, each of them gets a bamboo with a score greater than or equal to his/her lucky numb Er. 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 would 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

Idea: The prime number of the table, Euler function, two points.

Because to choose the smallest, so if ola[y]<ola[x] (x<y) then we have to choose X, so if the latter is less than the previous, we directly to the back of the update to the front so that the value of the Euler function is incremented, that does not use two points to choose the line.

#include <stdio.h>#include<algorithm>#include<iostream>#include<string.h>#include<stdlib.h>#include<math.h>#include<map>#include<Set>using namespacestd;BOOLprime[3300000+5];intsu[300000];typedefLong LongLl;typedefstructpp{intx; intID;} Ss;ss ola[3300000+5];intaa[10005];BOOLcmpstructPP nn,structpp mm) {        if(nn.x==mm.x)returnnn.id<mm.id; Else returnnn.x<mm.x;} typedef unsignedLong Longll;intMainvoid){        inti,j,k;  for(i=2; i<=7000; i++)        {                if(!Prime[i]) for(J=i; i*j<= (3300000); J + +) {Prime[i*j]=true; }        }        intans=0;  for(i=2; I<= (3300000); i++)                if(!Prime[i]) Su[ans++]=i;  for(i=1; i<=3300000; i++) {ola[i].id=i; ola[i].x=i; }         for(i=0; i<ans; i++)        {                 for(j=1; su[i]*j<=3300000; J + +) {Ola[su[i]*j].x=ola[su[i]*j].x/(Su[i]) * (su[i]-1); }        }         for(i=2; i<3300000; i++)        {                if(ola[i].x>ola[i+1].x) {ola[i+1].x=ola[i].x; }} scanf ("%d",&k); ints; intp,q;  for(s=1; s<=k; s++) {LL sum=0; scanf ("%d",&p);  for(i=0; i<p; i++) {scanf ("%d",&Aa[i]); intL=2; intR=4*1000000; intzz=0;  while(l<=r) {intMid= (l+r) >>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;}

1370-bi-shoe and Phi-shoe