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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#550487 | #9248. An Easy Math Problem | ucup-team3931# | TL | 0ms | 4060kb | C++14 | 1.1kb | 2024-09-07 13:09:33 | 2024-09-07 13:09:33 |
Judging History
This is the latest submission verdict.
- [2024-10-31 22:36:43]
- hack成功,自动添加数据
- (/hack/1098)
- [2024-10-31 22:13:58]
- hack成功,自动添加数据
- (/hack/1096)
- [2024-10-31 22:00:43]
- hack成功,自动添加数据
- (/hack/1095)
- [2024-09-07 13:09:33]
- Submitted
answer
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;
const int maxn = 2500;
ll n, d[maxn], m;
struct frac {
ll x, y;
frac(ll a = 0, ll b = 1) {
if (b < 0) {
a = -a;
b = -b;
}
x = a;
y = b;
}
};
inline bool operator < (const frac &a, const frac &b) {
return a.x * b.y < a.y * b.x;
}
void solve() {
m = 0;
scanf("%lld", &n);
for (ll i = 1; i * i <= n; ++i) {
if (n % i) {
continue;
}
d[++m] = i;
if (i * i != n) {
d[++m] = n / i;
}
}
set<frac> S;
sort(d + 1, d + m + 1);
for (int i = 1; i <= m; ++i) {
for (int j = i; j <= m; ++j) {
if (d[i] * d[j] > n) {
break;
}
if (n % (d[i] * d[j]) == 0) {
S.emplace(d[i], d[j]);
}
}
}
printf("%d\n", (int)S.size());
}
int main() {
int T = 1;
scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}
Details
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Test #1:
score: 100
Accepted
time: 0ms
memory: 4060kb
input:
10 1 2 3 4 5 6 7 8 9 10
output:
1 2 2 3 2 5 2 4 3 5
result:
ok 10 lines
Test #2:
score: -100
Time Limit Exceeded
input:
2000 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 6469693230 646969323...