Question Link
Enter a positive integer n to calculate the number of pairs of positive integers A <= B, which must be lcm (a, B) = n.
Idea: break down the quality factor and then obtain the logarithm directly by brute force.
Code:
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;typedef long long ll;const int MAXN = 1000;ll arr[MAXN];ll n;ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b);}int main() { while (scanf("%lld", &n) && n) { memset(arr, 0, sizeof(arr)); int cnt = 0; for (int i = 1; i <= sqrt(n); i++) { if (n % i == 0) { arr[cnt++] = i; arr[cnt++] = n / i; } } if (arr[cnt - 1] == arr[cnt - 2]) cnt--; int ans = 0; for (int i = 0; i < cnt; i++) { for (int j = i; j < cnt; j++) { if (((arr[i] * arr[j]) / gcd(arr[i], arr[j])) == n) { ans++; } } } printf("%lld %d\n", n, ans); } return 0;}
Uva10892-LCM cardinality (factorization factor)