[LightOJ](1236)Pairs Forming LCM ---- 唯一分解定理（质因数分解）

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 ≤ 1e14).

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

题意： 给你一个正整数n,让你求满足LCM（i,j） = n 的数对(i,j)有多少个

新知： 暴力不可解，然后做这个题时想了各种思路，都感觉时间复杂度爆炸。肝了一个多小时后选择放弃QAQ
然后一查题解发现了自己数论知识的盲区……

就是质因数分解和LCM和GCD之间有联系，自己发现一篇讲解不错，可以查看

https://www.cnblogs.com/linliu/p/5549544.html

扫描二维码关注公众号，回复： 2577219 查看本文章

AC代码:

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e7+1000;
const int MAXN = 1e3+5;
bool p[maxn];
int prime[maxn/10];
int a[MAXN];
int b[MAXN];
int tot;
int num;
void Init()
{
    tot = 0;
    memset(p,true,sizeof(p));
    p[0] = p[1] = false;
    for(LL i=2;i<maxn;i++)
    {
        if(p[i]){
            prime[tot++] = i;
            for(LL j=i*i;j<maxn;j+=i) p[j] = false;
        }
    }
}
void dec(LL x)
{
    num = 0;
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    for(int i=0;1LL*prime[i]*prime[i]<=x;i++)
    {
        if(x%prime[i] == 0){
            a[num] = prime[i];
            while(x%prime[i] == 0)
            {
                b[num]++;
                x/=prime[i];
            }
            num++;
        }
    }
    if(x!=1){
        a[num] = x;
        b[num] = 1;
        num++;
    }
}
int main()
{
    #ifdef LOCAL_FILE
    freopen("in.txt","r",stdin);
    #endif
//    ios_base::sync_with_stdio(false);
//    cin.tie(NULL),cout.tie(NULL);
    int t;
    int id = 1;
    scanf("%d",&t);
    Init();
    while(t--)
    {
        LL n;
        scanf("%lld",&n);
        dec(n);
        LL ans = 1;
        for(int i=0;i<num;i++)
            ans*=(2*b[i]+1);
        ans = ans/2+1;
        printf("Case %d: %lld\n",id++,ans);
    }
    return 0;
}