我相信这应该可以完成工作（来自@zch 的回答）：
#include <iostream>
#include <cmath>
auto MAX = [] (int A, int B) -> bool { return A > B ? A : B; };
auto MIN = [] (int A, int B) -> bool { return A < B ? A : B; };
using std::cout;
using std::cin;
int main() {
long long Z, M, N, T, low, high, temp, div;
int ans;
for (cin >> T; T--; ) {
cin >> Z >> M >> N;
temp = MIN(M, N);
low = MIN(sqrt(Z), temp);
high = MAX(M, N);
for( ans = 0; low > 0 && (Z / low) <= high; --low ) {
if ( Z % low == 0) {
++ans;
div = Z / low;
ans += (div != low && div <= temp);
}
//cout << temp << " * " << Z / temp << " = " << Z << "\n";
}
cout << ans << "\n";
}
return 0;
}
稍后会添加 cmets
使用 cmets 编写代码：
#include <iostream>
#include <cmath>
auto MAX = [] (int A, int B) -> bool { return A > B ? A : B; };
auto MIN = [] (int A, int B) -> bool { return A < B ? A : B; };
using std::cout;
using std::cin;
int main() {
long long Z, M, N, T, low, high, temp, div;
int ans;
for (cin >> T; T--; ) {
cin >> Z >> M >> N;
temp = MIN(M, N);
low = MIN(sqrt(Z), temp);//Lowest value <--We start iteration from this number
high = MAX(M, N); //Maximum value
for( ans = 0; low > 0 && (Z / low) <= high; --low ) {
//Number of things going on in this for-loop
//I will start by explaining the condition:
//We want to keep iterating until either low is below 1
// or when the expression (Z / low) > high.
//Notice that as the value of low approaches 0,
//the expression (Z / low) approaches inf
if ( Z % low == 0) {
//If this condition evaluates to true, we know 2 things:
/*Z is divisible by this value of low and
low is in the range of MIN(M,N) <--true*/
/*Because of our condition, (Z / low) is
within the range of MAX(M, N) <--true*/
++ans;
div = Z / low;
//This second part checks if the opposite is true i.e.
/*the value of low is in the range of
MAX(M, N) <--true*/
/*the value (Z / low) is in the range of
MIN(M, N) <--true only in some cases*/
ans += (div != low && div <= temp);
//(div != low) is to avoid double counting
/*An example of this is when Z, M, N have the values:
1000000, 1000000, 1000000
The value of low at the start is 1000 */
}
}
cout << ans << "\n";
}
return 0;
}