題目連結

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

對於每個素因子p，答案累加2*e+1，其中e是p的指數

#include<cstdio>
#include<cstring>
 

typedef long long ll;
const int N=1e7+100;
int t,p,cnt;
ll prime[N/10];
bool iscomp[N];
ll n,ans;
void primetable()
{
    p=0;
    for(int i=2;i<N;i++)
    {
        if(!iscomp[i])prime[p++]=i;
        for(int j=0;j<p&&prime[j]*i<N;j++)
        {
            iscomp[i*prime[j]]=1;
            if
 
(i%prime[j]==0)break;
        }
    }
}
void div()
{
    for(int i=0;i<p;i++)
    {
        if(prime[i]*prime[i]>n)break;
        cnt=0;
        while(n%prime[i]==0)
        {
            n/=prime[i];
            cnt++;
        }
        if(cnt)ans*=cnt*2+1;
    }
    if(n>1)ans*=3;
}
int main()
{
    //freopen("in.txt","r",stdin);
    scanf("%d",&t);
    primetable();
    int ca=0;
    while(t--)
    {
        ans=1;
        scanf("%lld",&n);
        div();
        printf("Case %d: %lld\n",++ca,ans/2+1);
    }
    return 0;
}

總結

無