牛客周赛52
小红的最大字典
思路:使用优先队列进行多路归并
#include<bits/stdc++.h>
#define int long long
#define x first
#define y second
#define endl '\n'
#define pq priority_queue
using namespace std;
typedef pair<int,int> pii;
void solve(){
priority_queue<pair<char,string>>q;
int n;
for(int i = 0;i < n;i ++){
string s;
cin >> s;
q.push({s[0], s});
}
string ans = "";
while(q.size()){
auto [c, s] = q.top();
q.pop();
s.erase(0, 1);
ans += c;
if(s.size() > 0)
q.push({s[0], s});
}
cout << ans << endl;
}
signed main()
{
ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int t = 1;
while(t--)
{
solve();
}
return 0;
}
思路:这里有个结论,只要左右括号相等,通过循环移位一定可以成为合法的字符串序列,所以这道题就变成一道组合数学的题了
int fac[N], inf[N];
int ksm(int a, int b) {
int res = 1;
while (b) {
if (b & 1) res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int C(int n, int m) {
return fac[n] * inf[n - m] % mod * inf[m] % mod;
}
void solve() {
fac[0] = inf[0] = 1;
for (int i = 1; i < N; i++) {
fac[i] = fac[i - 1] * i % mod;
inf[i] = inf[i - 1] * ksm(i, mod - 2) % mod;
}
int n; cin >> n;
string s; cin >> s;
int c1 = 0, c2 = 0, cv = 0;
for (auto i : s) {
c1 += (i == '(');
c2 += (i == ')');
cv += (i == '?');
}
if (c1 > n / 2 || c2 > n / 2 || n & 1) {
cout << 0 << '\n';
} else cout << C(cv, abs(c1 - c2) + (cv - abs(c1 - c2)) / 2) << '\n';
}