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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 |
---|---|---|---|---|---|---|---|---|---|
#505995 | #6429. Let's Play Curling | Grunray | WA | 84ms | 5196kb | C++20 | 3.1kb | 2024-08-05 14:29:55 | 2024-08-05 14:29:56 |
Judging History
answer
#define _CRT_SECURE_NO_WARNINGS
#define itn int
#define PII pair<int, int>
#define PLI pair<long long, int>
#define fep(i, a, b) for(int i = (a); i >= (b); --i)
#define rep(i, a, b) for(int i = (a); i <= (b); ++i)
#include<bits/stdc++.h>
#include<unordered_map>
using ll = long long;
using ldou = long double;
using unll = unsigned long long;
using namespace std;
inline int read() {
int x = 0, f = 1;
char ch = getchar();
while (!isdigit(ch)) { f = ch != '-'; ch = getchar(); }
while (isdigit(ch)) { x = (x << 1) + (x << 3) + (ch ^ 48); ch = getchar(); }
return f ? x : -x;
}
ll gcd(ll a, ll b) { // 最大公约数
while (b ^= a ^= b ^= a %= b)
;
return a;
}
ll lcm(ll a, ll b) { // 最小公倍数
return a / gcd(a, b) * b;
}
ll qmi(ll m, ll k, ll p) { // 快速幂
//求 m^k mod p,时间复杂度 O(logk)。
//m为底数,k为幂
ll res = 1 % p, t = m;
while (k) {
if (k & 1) res = res * t % p;
t = t * t % p;
k >>= 1;
}
return res;
}
unll qmi(unll m, unll k, unll p) { //龟速乘
ll res = 0, t = m;
while (k) {
if (k & 1) res = (res + t) % p;
k >>= 1;
t = (t << 1) % p;
}
return res;
}
////////////////////////////////////////////////////////////////////////////////
const int N = 1e5 + 50;
const int M = 2e5 + 50;
const ll INF = 0x3f3f3f3f;
const ll MODE = ll(998244353);
const double Pi = 3.1415926;
const double eps = 1e-8;
const int dx[4] = { 1,-1, 0, 0 };
const int dy[4] = { 0, 0, 1,-1 };
//priority_queue<int> p;//这是一个大根堆q
//priority_queue<int, vector<int>, greater<int> >q;//这是一个小根堆q
//priority_queue<ll, vector<ll>, greater<ll> >pq; // 小根
ll n, m;
string str;
ll a[N], b[N];
set<ll>sa;
set<ll>sb;
void solve() {
cin >> n >> m;
/*rep(i, 1, n) cin >> a[i];
rep(j, 1, m) cin >> b[j];*/
ll x;
rep(i, 1, n) {
//cin >> x;
cin >> a[i];
//sa.insert(x);
}
rep(i, 1, m) {
//cin >> x;
//sb.insert(x);
cin >> b[i];
}
set<ll>::iterator it;
int pos = 1;
/*for (it = sa.begin(); it != sa.end(); it++) {
a[pos++] = (*it);
}
n = pos - 1;*/
sort(a + 1, a + n + 1);
sort(b + 1, b + n + 1);
/*pos = 1;
for (it = sb.begin(); it != sb.end(); it++) {
b[pos++] = (*it);
}
m = pos;*/
m = ll(unique(b + 1, b + m + 1) - b) - 1;
/*cout << " a ";
for (int i = 1; i <= n; i++)cout << a[i] << ' ';
cout << '\n';
cout << " b ";
for (int i = 1; i <= m; i++)cout << b[i] << ' ';
cout << '\n';*/
b[0] = 0;
b[++m] = INF;
ll res = 0;
ll l, r;
ll mid;
rep(i, 1, m) {
l = ll(upper_bound(a + 1, a + n + 1, b[i - 1]) - a);
r = ll(lower_bound(a + 1, a + n + 1, b[i]) - a) - 1;
mid = r - l + 1;
res = max(res, mid);
}
//cout << "****** ";
if (!res) cout << "Impossible\n";
else cout << res << '\n';
/*sa.clear();
sb.clear();*/
}
signed main() {
std::ios::sync_with_stdio(false); std::cin.tie(0), std::cout.tie(0);
/*freopen("out.txt", "r", stdin);
freopen("wrt.txt", "w", stdout);*/
int TTT = 1; cin >> TTT;
while (TTT--) {
solve();
}
/*while (cin >> n >> m) {
solve();
}*/
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 3716kb
input:
3 2 2 2 3 1 4 6 5 2 5 3 7 1 7 3 4 3 1 10 1 1 7 7
output:
2 3 Impossible
result:
ok 3 lines
Test #2:
score: -100
Wrong Answer
time: 84ms
memory: 5196kb
input:
5553 12 19 8 8 11 18 12 9 15 38 6 32 30 30 17 28 33 2 37 20 11 38 36 18 18 30 20 33 13 31 33 37 8 12 6 7 12 14 2 19 2 17 7 4 20 1 13 7 18 23 22 1 16 8 7 5 2 4 2 4 5 8 12 13 16 6 6 5 16 11 5 7 5 13 3 8 3 11 6 9 11 13 8 11 17 19 944782509 244117333 140979583 661724696 617847780 321687699 418677763 725...
output:
9 3 4 3 8 8 11 4 2 4 11 5 4 3 15 3 9 4 6 13 6 4 5 4 12 2 6 9 3 3 1 4 6 11 4 3 4 5 4 6 9 4 5 13 3 5 4 5 5 1 6 10 4 5 5 5 6 8 3 4 10 2 4 5 4 5 6 10 2 5 5 4 6 5 1 7 1 3 2 4 4 12 17 2 2 3 19 5 5 4 3 4 5 4 4 3 8 5 13 2 3 6 9 6 19 2 1 2 4 7 7 10 7 12 1 2 2 4 6 10 1 3 4 4 13 2 1 13 4 8 14 6 2 13 6 3 6 5 6 ...
result:
wrong answer 1st lines differ - expected: '1', found: '9'