LightOJ 1236 Pairs Forming LCM(唯一分解定理+素数刷选)

Pairs Forming LCM

Description

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

题解：把n分解，记下各个因子的个数，假设（a,b）满足LCM（a,b）==n，
则对于n的每种因子num[i],a,b中至少有一个数该种因子的个数为num[i];
另外一个则有k种情况，还有一种情况是a=b,即两个数该种因子都为num[i].
所以，一种因子的情况为C(2,1)*num[i]+1.将每种因子的情况数相乘就是答案了。

#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>

#define ll long long
#define N 10000010

using namespace std;

ll n;
vector<int>vec;
bool _is
;

void init() {
    memset(_is,0,sizeof _is);
    _is[0]=_is[1]=1;
    for(int i=2; i<N; i++) {
        if(!_is[i]) {
            vec.push_back(i);
            for(int j=i+i; j<N; j+=i)
                _is[j]=1;
        }
    }
}

int main() {
    freopen("test.in","r",stdin);
    init();
    int t;
    int ca=1;
    cin>>t;
    while(t--) {
        scanf("%lld",&n);
        ll ans=1;
        int k;
        for(int i=0; i<vec.size(); i++) {
            if(n<vec[i])break;
            k=0;
            if(n%vec[i]==0) {
                while(n%vec[i]==0) {
                    k++;
                    n/=vec[i];
                }
            }
            if(k)
                ans*=(2*(k)+1);
        }
        if(n!=1)ans*=3;//最后一个因子
        printf("Case %d: %lld\n",ca++,(ans+1)/2);//由题意，答案要除以二
    }
    return 0;
}