LightOJ

Find the result of the following code:

long long pairsFormLCM( int n ) {
    long long res = 0;
    for( int i = 1; i <= n; i++ )
        for( int j = i; j <= n; j++ )
           if( lcm(i, j) == n ) res++; // lcm means least common multiple
    return res;
}

A straight forward implementation of the code may time out. If you analyze the code, you will find that the code actually counts the number of pairs(i, j) for which lcm(i, j) = n and (i ≤ j).

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 10¹⁴).

Output

For each case, print the case number and the value returned by the function 'pairsFormLCM(n)'.

Sample Input

15

2

3

4

6

8

10

12

15

18

20

21

24

25

27

29

Sample Output

Case 1: 2

Case 2: 2

Case 3: 3

Case 4: 5

Case 5: 4

Case 6: 5

Case 7: 8

Case 8: 5

Case 9: 8

Case 10: 8

Case 11: 5

Case 12: 11

Case 13: 3

Case 14: 4

Case 15: 2

这个题我一开始怕不是想到天边去了………………

设n=p1^e1*p2^e2*p3^e3*.....^pn^en.

则，当a，b的LCM(a,b)==n时

令a=p1^a1*p2^a2*...*pn^an

b=p1^b1*p2^b2*...*pn^bn

（ai，bi可为0）

则有max（a1，b1）=e1……max（ai，bi）=ei……max（an，bn）=en

这样的ai，bi取值的方案数为s=（2*（ei+1）-1）种，然后乘起来得到ans，然后ans=ans/2+1，加1应该是因为LCM（n，n）不会重复（应该吧……）

#include <iostream>#include <stdio.h>#include <math.h>#include <string.h>#include <vector>#include <algorithm>using namespace std;typedef long long LL;const int MAXN=1e7+9;#define fi first#define se secondint prime[700000-23],total;bool isprime[MAXN];void make(){    int m=MAXN-3;    memset(isprime,true,sizeof(isprime));    isprime[0]=isprime[1]=false;    total=0;    for(int i=2;i<=m;i++)    {        if(isprime[i])prime[++total]=i;        for(int j=1;j<=total&&prime[j]*i<=m;j++)        {            isprime[prime[j]*i]=false;            if(i%prime[j]==0)break;        }    }    memset(isprime,false,sizeof(isprime));    for(int i=1;i<=total;i++)isprime[prime[i]]=true;}int main(){    make();    int T,cas=0;    //cout<<total;    scanf("%d",&T);    while(T--)    {        LL n;        scanf("%lld",&n);        LL ans=1;        for(int i=1;i<=total&&prime[i]*prime[i]<=n;i++)        {            LL s=0;            if(n%prime[i]==0)            {                while(n%prime[i]==0)                {                    n/=prime[i];                    s++;                }            }            ans*=(s+1)*2-1;        }        if(n>1)ans*=3;        printf("Case %d: %lld\n",++cas,ans/2+1);    }    return 0;}