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技术 2022年11月23日
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目录

Contest Info


[Practice Link](https://codeforces.com/contest/1251)

Solved A B C D E1 E2 F
6/7 O O O O O O
  • O 在比赛中通过
  • Ø 赛后通过
  • ! 尝试了但是失败了
  • – 没有尝试

Solutions


A. Broken Keyboard

题意:

有一个打字机,如果某个按键是好的,那么按下那个按键之后会在打字槽中追加一个该字符,如果是坏的则会追加两个。

现在给出打印槽中最后的结果,问有哪些按键能确定一定是好的。

思路:

将连续的相同字符取出来,如果长度为奇数,那么一定能确定该字符对应的按键是好的。

代码:

view code
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
#define fi first
#define se second
#define endl "\n"
using namespace std;
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair <int, int>;
using pLL = pair <ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }
template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }
template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }
inline int rd() { int x; cin >> x; return x; }
template <class T> inline void rd(T &x) { cin >> x; }
template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }
#define dbg(x...) do { cout << "\033[32;1m" << #x << " -> "; err(x); } while (0)
void err() { cout << "\033[39;0m" << endl; }
template <class T, class... Ts> void err(const T& arg, const Ts&... args) { cout << arg << ' '; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) { for (auto &v : arg) cout << v << ' '; err(args...); }
inline void pt() { cout << endl; }
template <class T, class... Ts> void pt(const T& arg, const Ts&... args) { cout << arg << ' '; pt(args...); }
template <template<typename...> class T, typename t, typename... A>
void pt(const T <t> &arg, const A&... args) { for (auto &v : arg) cout << v << ' '; pt(args...); }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }
//head
constexpr int N = 1e5 + 10;
int n, cnt[30]; string s;
void run() {
cin >> s;
memset(cnt, -1, sizeof cnt);
for (int i = 0, len = s.size(), num = 0; i <= len; ++i) {
if (i == len) {
if (num & 1) {
cnt[s[i - 1] - 'a'] = 1;
}
} else if (i && s[i] != s[i - 1]) {
if (num & 1) {
cnt[s[i - 1] - 'a'] = 1;
}
num = 0;
}
++num;
}
for (int i = 0; i < 26; ++i) if (cnt[i] > 0)
cout << char(i + 'a');
cout << endl;}int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
cout << fixed << setprecision(20);
int _T = rd();
while (_T--) run();
return 0;
}

B. Binary Palindromes

题意:

给出\(n\)个\(01\)串,可以任意交换任意两个字符串的任意两个位置的字符,问最终最多能有多少回文串。

思路:

任意交换,只需要考虑\(0\)有多少个,\(1\)有多少个,然后根据原串长度贪心构造。

代码:

view code
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
#define fi first
#define se second
#define endl "\n"
using namespace std;
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair <int, int>;
using pLL = pair <ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }
template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }
template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }
inline int rd() { int x; cin >> x; return x; }
template <class T> inline void rd(T &x) { cin >> x; }
template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }
#define dbg(x...) do { cout << "\033[32;1m" << #x << " -> "; err(x); } while (0)
void err() { cout << "\033[39;0m" << endl; }
template <class T, class... Ts> void err(const T& arg, const Ts&... args) { cout << arg << ' '; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) { for (auto &v : arg) cout << v << ' '; err(args...); }
inline void pt() { cout << endl; }
template <class T, class... Ts> void pt(const T& arg, const Ts&... args) { cout << arg << ' '; pt(args...); }
template <template<typename...> class T, typename t, typename... A>
void pt(const T <t> &arg, const A&... args) { for (auto &v : arg) cout << v << ' '; pt(args...); }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }
//head
constexpr int N = 1e5 + 10;
int n, cnt[2], len[110];
void run() {
n = rd();
cnt[0] = cnt[1] = 0;
for (int i = 1; i <= n; ++i) {
string s; cin >> s;
len[i] = s.size();
for (auto &c : s) ++cnt[c - '0'];
}
sort(len + 1, len + 1 + n);
int res = 0;
for (int i = 1; i <= n; ++i) {
if (len[i] & 1) {
if (cnt[0] & 1) {
--cnt[0];
} else if (cnt[1] & 1) {
--cnt[1];
} else if (cnt[0]) {
--cnt[0];
} else if (cnt[1]) {
--cnt[1];
} else {
break;
}
--len[i];
}
if (len[i] > cnt[0]) {
if (cnt[0] & 1) {
len[i] = len[i] - cnt[0] + 1;
cnt[0] = 1;
} else {
len[i] -= cnt[0];
cnt[0] = 0;
}
} else {
cnt[0] -= len[i];
len[i] = 0;
}
if (len[i] > cnt[1]) {
if (cnt[1] & 1) {
len[i] = len[i] - cnt[1] + 1;
cnt[1] = 1;
} else {
len[i] -= cnt[1];
cnt[1] = 0;
}
} else {
cnt[1] -= len[i];
len[i] = 0;
}
if (len[i]) break;
++res;
}
pt(res);
}int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
cout << fixed << setprecision(20);
int _T = rd();
while (_T--) run();
return 0;
}

C. Minimize The Integer

题意:

给出一个\(n\)位的整数,可能有前导\(0\),现在可以任意交换两个相邻的并且奇偶性不同的数的位置,最终结果也可以存在前导\(0\),问最终结果的最小值是多少。

思路:

显然,所有奇偶性相同的数的相对位置不变,那么将奇数取出来,偶数取出来,然后贪心取头即可。

代码:

view code
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
#define fi first
#define se second
#define endl "\n"
using namespace std;
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair <int, int>;
using pLL = pair <ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }
template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }
template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }
inline int rd() { int x; cin >> x; return x; }
template <class T> inline void rd(T &x) { cin >> x; }
template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }
#define dbg(x...) do { cout << "\033[32;1m" << #x << " -> "; err(x); } while (0)
void err() { cout << "\033[39;0m" << endl; }
template <class T, class... Ts> void err(const T& arg, const Ts&... args) { cout << arg << ' '; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) { for (auto &v : arg) cout << v << ' '; err(args...); }
inline void pt() { cout << endl; }
template <class T, class... Ts> void pt(const T& arg, const Ts&... args) { cout << arg << ' '; pt(args...); }
template <template<typename...> class T, typename t, typename... A>
void pt(const T <t> &arg, const A&... args) { for (auto &v : arg) cout << v << ' '; pt(args...); }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }
//head
constexpr int N = 3e5 + 10;
int n; char s[N];
void out(vector <int> &vec) {
cout << vec.back();
vec.pop_back();
}
void run() {
vector <int> vec[2];
cin >> (s + 1);
for (int i = 1; s[i]; ++i) {
int num = s[i] - '0';
vec[num & 1].push_back(num);
}
reverse(vec[0].begin(), vec[0].end());
reverse(vec[1].begin(), vec[1].end());
while (!vec[0].empty() || !vec[1].empty()) {
if (vec[0].empty()) {
out(vec[1]);
} else if (vec[1].empty()) {
out(vec[0]);
} else {
if (vec[0].back() < vec[1].back()) {
out(vec[0]);
} else {
out(vec[1]);
}
}
}
cout << endl;
}int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
cout << fixed << setprecision(20);
int _T; cin >> _T;
while (_T--) run();
return 0;
}

D. Salary Changing

题意:

有\(n\)个人,你是老板,你手里有\(s\)元钱,要给这\(n\)个人发工资,每个人工资的范围是\([l_i, r_i]\),要如何发工资使得你的钱够用并且\(n\)个人工资的中位数最高。

思路:

首先给每个人发\(l_i\)工资,那么得到一个答案的下界,那么发现这个下界到\(INF\)这个范围,答案具有单调性,二分然后贪心\(check\)即可。

代码:

view code
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
#define fi first
#define se second
#define endl "\n"
using namespace std;
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair <int, int>;
using pLL = pair <ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }
template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }
template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }
inline int rd() { int x; cin >> x; return x; }
template <class T> inline void rd(T &x) { cin >> x; }
template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }
#define dbg(x...) do { cout << "\033[32;1m" << #x << " -> "; err(x); } while (0)
void err() { cout << "\033[39;0m" << endl; }
template <class T, class... Ts> void err(const T& arg, const Ts&... args) { cout << arg << ' '; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) { for (auto &v : arg) cout << v << ' '; err(args...); }
inline void pt() { cout << endl; }
template <class T, class... Ts> void pt(const T& arg, const Ts&... args) { cout << arg << ' '; pt(args...); }
template <template<typename...> class T, typename t, typename... A>
void pt(const T <t> &arg, const A&... args) { for (auto &v : arg) cout << v << ' '; pt(args...); }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }
//head
constexpr int N = 2e5 + 10;
int n; ll s;
pII a[N];
bool check(ll x) {
int l = 0, r = 0;
ll remind = s;
for (int i = 1; i <= n; ++i) {
if (a[i].se < x) {
remind -= a[i].fi;
++l;
} else if (a[i].fi > x) {
remind -= a[i].fi;
++r;
}
}
if (l > n / 2 || r > n / 2) return false;
for (int i = 1; i <= n; ++i) {
if (a[i].fi <= x && a[i].se >= x) {
if (l < n / 2) {
++l;
remind -= a[i].fi;
} else {
remind -= x;
}
}
}
return remind >= 0;
}
void run() {
cin >> n >> s;
for (int i = 1; i <= n; ++i) a[i].fi = rd(), a[i].se = rd();
sort(a + 1, a + 1 + n);
ll l = a[n / 2 + 1].fi, r = 1e9, res = l;
while (r - l >= 0) {
ll mid = (l + r) >> 1;
if (check(mid)) {
res = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
pt(res);
}int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
cout << fixed << setprecision(20);
int _T = rd();
while (_T--) run();
return 0;
}

E2. Voting (Hard Version)

题意:

有\(n\)个人,你可以花费\(p_i\)让第\(i\)个人投票,或者拉够\(m_i\)个人为你投票,这个人就会为你投票。

现在你想让所有人都为你投票,需要花费的最小代价是多少?

思路:

假设我知道有\(x\)个人不需要花费代价能让他们为我免费投票,那么我假设刚开始对每个人都付了钱让他们投票,现在就是要去除\(x\)个人的花费,使得去除的花费最大。

那么我们从\(i \in [x, n]\)扫一遍,用一个大根堆维护\(m_i < i\)的所有人的最大\(p_i\),每次取出堆顶的\(p_i\)即可。

并且容易发现\(x\)具有单调性,直接二分即可。

代码:

view code
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
#define fi first
#define se second
#define endl "\n"
using namespace std;
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair <int, int>;
using pLL = pair <ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2> inline void chadd(T1 &x, T2 y) { x += y; while (x >= mod) x -= mod; while (x < 0) x += mod; }
template <class T1, class T2> inline void chmax(T1 &x, T2 y) { if (x < y) x = y; }
template <class T1, class T2> inline void chmin(T1 &x, T2 y) { if (x > y) x = y; }
inline int rd() { int x; cin >> x; return x; }
template <class T> inline void rd(T &x) { cin >> x; }
template <class T> inline void rd(vector <T> &vec) { for (auto &it : vec) cin >> it; }
#define dbg(x...) do { cout << "\033[32;1m" << #x << " -> "; err(x); } while (0)
void err() { cout << "\033[39;0m" << endl; }
template <class T, class... Ts> void err(const T& arg, const Ts&... args) { cout << arg << ' '; err(args...); }
template <template<typename...> class T, typename t, typename... A>
void err(const T <t> &arg, const A&... args) { for (auto &v : arg) cout << v << ' '; err(args...); }
inline void pt() { cout << endl; }
template <class T, class... Ts> void pt(const T& arg, const Ts&... args) { cout << arg << ' '; pt(args...); }
template <template<typename...> class T, typename t, typename... A>
void pt(const T <t> &arg, const A&... args) { for (auto &v : arg) cout << v << ' '; pt(args...); }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline ll qpow(ll base, ll n) { ll res = 1; while (n) { if (n & 1) res = res * base % mod; base = base * base % mod; n >>= 1; } return res; }
//head
constexpr int N = 2e5 + 10;
int n; vector <vector<int>> vec;
ll gao(int x) {
ll tot = 0;
priority_queue <int, vector<int>, less<int>> pq;
for (int i = 1; i <= n; ++i) {
for (auto &it : vec[i]) pq.push(it);
if (i >= x) {
if (pq.empty()) return 0;
tot += pq.top(); pq.pop();
}
}
return tot;
}
void run() {
n = rd();
vec.clear(); vec.resize(n + 1);
ll tot = 0;
for (int i = 1, m, p; i <= n; ++i) {
m = rd(); p = rd();
vec[m + 1].push_back(p);
tot += p;
}
int l = 1, r = n; ll Max = 0;
while (r - l >= 0) {
int mid = (l + r) >> 1;
ll tmp = gao(mid);
chmax(Max, tmp);
//dbg(l, r, mid, Max);
if (tmp > 0) {
r = mid - 1;
} else {
l = mid + 1;
}
}
pt(tot - Max);
}int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
cout << fixed << setprecision(20);
int _T = rd();
while (_T--) run();
return 0;
}
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