把n进行质因数分解,设n的某个质因子为p,其指数为cnt.
对i和j包含的相同质因子p,设指数分别为a,b,由于LCM(i,j)=n,则有cnt=max(a,b).
a,b共有2*(cnt+1)-1种方案(乘以2是a,b互换,减1是剔除a=b的重复情况)
之后由乘法原理,得到了(i,j)的方案数res, 但其中包括了i>j的部分.i<j 和 i>j是对称的, 则 (res+1)/2就是题目要求的了.
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>
#include <map>
#include <set>
#include <string>
#include <vector>
#define ll long long
using namespace std;
const int maxn = 1e7;
bool v[maxn + 1];
vector<int> prime;
void primes() {
v[0] = v[1] = 1;
for (int i = 2; i <= maxn; ++i) {
if (!v[i]) {
prime.push_back(i);
for (int j = i; j <= maxn / i; ++j) {
v[i * j] = 1;
}
}
}
}
ll divide(ll n) {
ll res = 1;
for (int i = 0; i < prime.size(); ++i) {
if (prime[i] > n) break;
int t = 0;
if (n % prime[i] == 0) {
while (n % prime[i] == 0) {
t++;
n /= prime[i];
}
res *= 2 * (t + 1) - 1;
}
}
if (n > 1) res *= 3;
return (res + 1) / 2;
}
int main() {
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int T, kase = 0;
primes();
cin >> T;
while (T--) {
long long n;
cin >> n;
printf("Case %d: %lld\n", ++kase, divide(n));
}
return 0;
}