LCM Cardinality
Input: Standard Input
Output: Standard Output
Time Limit: 2 Seconds
A pair of numbers has a unique LCM but a single number can be the LCM of more than one possible pairs. For example 12 is the LCM of (1, 12), (2, 12), (3,4) etc. For a given positive integer N, the number of different integer pairs with LCM is equal to N can be called the LCMcardinality of that number N. In this problem your job is to find out the LCM cardinality of a number.
Input
The input file contains at most 101 lines of inputs. Each line contains an integer N (0<N<=2*109). Input is terminated by a line containing a single zero. This line should not be processed.
Output
For each line of input except the last one produce one line of output. This line contains two integers N and C. Here N is the input number and Cis its cardinality. These two numbers are separated by a single space.
Sample Input Output for Sample Input
2 12 24 101101291 0 |
2 2 12 8 24 11 10110129 |
题意：给出a和b的最小公倍数N。找出符合条件的a、b有多少对。
分析：1. 设n = LCM(a,b) = (p1^r1) * (p2^r2) * (p3^r3) … (pm^rm)
又设a=(p1^a1) * (p2^a2) * (p3^a3) … (pm^am),
b=(p1^b1) * (p2^b2) * (p3^b3)… (pm^bm)
由LCM的定义有ri = max{ai, bi}
所以对于每一个ri，ai和bi中至少有一个要取ri
2. 对于ai取ri的情况，bi能够取[0,ri-1]的随意整数，这有ri种情况；
bi取ri的情况相同是ri种 。
最后加上ai和bi都取ri的情况，共同拥有(2*ri+1)种情况
3. 由于这么考虑把(a,b)和(b,a)算反复了，但(n,n)的情况仅仅算了一遍。所以最后要ans= (ans+1)/2=ans/2+1（由于ans是奇数）
4. 优化：仅仅考虑√n范围内的质数，但这样会存在漏掉一个大质数的情况（比方n=2*101） 。这个大质数的幂次仅仅能为1（即少算了一个*(2*1+1)），所以在这样的情况发生时要补上ans*=3，写成 位运算就是ans+=ans<<1。
#include <cstdio>
#include <cmath>
int n;
void get_ans() {
int tmp = n;
int m = (int)sqrt(n + 0.5);
long long ans = 1;
for(int i = 2; i <= m; i += 2) {
if(n % i == 0) {
int cnt = 0;
while(n % i == 0) {
n /= i;
cnt++;
}
ans *= (cnt << 1) + 1;
}
if(i == 2) i--;
}
if(n > 1) ans += (ans<<1);
ans = (ans >> 1) + 1;
printf("%d %lld
", tmp, ans);
}
int main() {
while(~scanf("%d", &n) && n) {
get_ans();
}
return 0;
}