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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#103479 | #5661. Multi-Ladders | cardinal_city# | AC ✓ | 2ms | 3396kb | C++17 | 1.2kb | 2023-05-06 03:56:40 | 2023-05-06 03:56:41 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int, int> pii;
#define all(x) (x).begin(), (x).end()
#define sz(x) (int)x.size();
#define rep(i,j,k) for (int i = j; i < (k); i++)
#define in(mp, v) (mp.find(v) != mp.end())
#define smx(a, b) a = max(a, b)
#define smn(a, b) a = min(a, b)
#define pb push_back
#define endl '\n'
const ll MOD = 1e9 + 7;
const ld EPS = 1e-9;
ll modpow(ll b, ll e) {
ll ans = 1;
for (; e; b = b * b % MOD, e /= 2)
if (e & 1) ans = ans * b % MOD;
return ans;
}
int main() {
cin.tie(0)->sync_with_stdio(0);
int t; cin >> t;
while (t--) {
ll n, k, l; cin >> n >> k >> l;
if (l < 2 || (k % 2 == 1 && l < 3)) {
cout << 0 << "\n";
continue;
}
ll p1 = modpow(l - 1, k);
p1 += (k % 2 == 1 ? -1 : 1) * (l - 1);
while (p1 < 0) {
p1 += MOD;
}
p1 %= MOD;
ll p = (l * l - 3 * l + 3 + 5 * MOD) % MOD;
ll p2 = modpow(p, (n - 1) * k);
cout << p1 * p2 % MOD << "\n";
}
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 3396kb
input:
1 2 3 3
output:
162
result:
ok single line: '162'
Test #2:
score: 0
Accepted
time: 2ms
memory: 3316kb
input:
20 2 3 3 1 3 3 10 3 0 10 3 2 1 21 2 1 22 0 2000 15000 2000 12000 30000 200000 1000000000 3 3 2 1000000000 3 2 3 100000000 1000000000 1000000000 10 1000000000 3 100000000 2 1000000000 100000000 1 1000000000 10 1 1000000000 100000000 1 1000 100000000 1000000000 1000000000 0 1000000000 1000000000 1 100...
output:
162 6 0 0 0 0 349400141 243010659 52489881 53690844 176686901 218103365 558243892 991895211 693053429 883715672 80402569 0 0 311752813
result:
ok 20 lines