LightOJ 1236 Pairs Forming LCM （唯一分解定理）

Pairs Forming LCM Time Limit:2000MS    
Memory Limit:32768KB     64bit IO Format:%lld & %llu

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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

思路：把n分解成素因数的形式n=p1^c1*p2^c2*...pm^cm

假设已找到一对（a，b）的lcm=n

有a=p1^d1*p2^d2*...pm^dm

b=p1^e1*p2^e2*...pm^em

易知max(di,ei)=ci

所以必有一个di或ei = ci；

种数为 （2 * （ci + 1） - 1），都为ci是重复了，所以减一；

所以有序对(a,b)的种数ans=(2*c1+1)*(2*c2+1)*(2*c3+1)*...*(2*cm+1)

但是要求求无序对的种数，已知(a,b)(b,a)重复，除了（n，n）只算了一个之外，所以ans = ans/2 +1

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
long long a, c;
long long prime[700000];
bool vis[10000005];//用bool定义减少内存
int tot;
void isprime() // 线性素数筛选
{
tot = 0;
for(int i = 2; i <= 10000000; ++ i)
{
if(!vis[i])
{
prime[tot ++] = i;
}
for(int j = 0; j < tot && prime[j] * i < 10000000; ++ j)
{
vis[prime[j] * i] = 1;
if(i % prime[j] == 0)
break;
}
}
}
//char b[1000005];
int main()
{
int t;
long long n;
/*printf("%d\n", 32768 * 1024 / sizeof(double));
printf("%d\n", 32768 * 1024 / sizeof(bool));
printf("%d\n", 32768 * 1024 / sizeof(long long));*/
scanf("%d", &t);
int tt = 1;
isprime();
while(t --)
{
scanf("%lld", &n);
long long ans = 1;
for(int i = 0; i < tot; ++ i)
{
if(prime[i] * prime[i] > n)
break;
long long c = 0;
while(n % prime[i] == 0)
{
c ++;
n /= prime[i];
}
if(c)
ans *= (2 * c + 1);
}
if(n > 1)
ans *= 3;
printf("Case %d: %lld\n", tt ++, ans / 2 + 1);
}

}