传送门
A There Are Two Types Of Burgers
简单的枚举每一种情况，来贪心的选择
//
// Author : north_h
// File : A.cpp
// Time : 2023/7/24/11:59
// _ _ _
// _ __ ___ _ __| |_| |__ | |__
//| '_ \ / _ \| '__| __| '_ \ | '_ \
//| | | | (_) | | | |_| | | | | | | |
//|_| |_|\___/|_| \__|_| |_|___|_| |_|
// |_____|
#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define met_x(a) memset(a,0x3f,sizeof a)
#define mpy(a, b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define fi first
#define se second
#define PII pair<int,int>
#define PDD pair<double,double>
#define PCI pair<char,int>
#define ALL(a) a.begin(),a.end()
#define rALL(a) a.begin(),a.end()
#define int128 __int128
#define endl '\n'
const int N = 10010;
const int M = 110;
const int MOD = 998244353;
const int EPS = 1e-8;
const int INF = 0x3f3f3f3f;
using namespace std;
void solve() {
int b, p, f, h, c;
cin >> b >> p >> f >> h >> c;
int ans = 0;
for (int i = 0; (i * 2) <= b && i <= p; i++) {
int x = b - i * 2;
int val = i * h + min(x / 2, f) * c;
ans=max(ans,val);
}
cout << ans << endl;
}
int main() {
IOS;
int h_h = 1;
cin >> h_h;
while (h_h--)solve();
return 0;
}
B Square Filling
这个题没有规定次数，直接枚举整个矩阵的位置，再2x2的方格里，如果全部都为1，就标记为true，最后验证vis矩阵是否和所给的矩阵相同即可，这个题因为空间不小心开大了还MLE了一发，不好说
//
// Author : north_h
// File : B.cpp
// Time : 2023/7/24/12:06
// _ _ _
// _ __ ___ _ __| |_| |__ | |__
//| '_ \ / _ \| '__| __| '_ \ | '_ \
//| | | | (_) | | | |_| | | | | | | |
//|_| |_|\___/|_| \__|_| |_|___|_| |_|
// |_____|
#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define met_x(a) memset(a,0x3f,sizeof a)
#define mpy(a, b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define fi first
#define se second
#define PII pair<int,int>
#define PDD pair<double,double>
#define PCI pair<char,int>
#define ALL(a) a.begin(),a.end()
#define rALL(a) a.begin(),a.end()
#define int128 __int128
#define endl '\n'
const int N = 100;
const int M = 110;
const int MOD = 998244353;
const int EPS = 1e-8;
const int INF = 0x3f3f3f3f;
using namespace std;
int n,m;
int g[N][N];
int vis[N][N];
void solve() {
cin >> n >> m;
int cnt = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> g[i][j];
if (g[i][j] == 0)cnt++;
}
}
if (cnt == n * m) {
cout << 0 << endl;
return;
}
vector <PII> ans;
for (int i = 0; i < n - 1; i++) {
for (int j = 0; j < m - 1; j++) {
if (g[i][j] == 1 && g[i][j + 1] == 1 && g[i + 1][j] == 1 && g[i + 1][j + 1] == 1) {
ans.push_back({i + 1, j + 1});
vis[i][j] = vis[i][j + 1] = vis[i + 1][j] = vis[i + 1][j + 1] = 1;
}
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (vis[i][j] != g[i][j]) {
cout << -1 << endl;
return;
}
}
}
cout << ans.size() << endl;
for (auto i: ans)cout << i.fi << ' ' << i.se << endl;
}
int main() {
IOS;
int h_h = 1;
//cin >> h_h;
while (h_h--)solve();
return 0;
}
C Gas Pipeline
dp，只有两个状态，要么用高的支柱，要么用低的支柱，因为最开始只能为低支柱，送总第一个位置开始推后面的状态，要注意一个特殊情况，如果这个位置是高支柱，那么前面一定是高支柱，最后也一定是低支柱
//
// Author : north_h
// File : C.cpp
// Time : 2023/7/24/12:26
// _ _ _
// _ __ ___ _ __| |_| |__ | |__
//| '_ \ / _ \| '__| __| '_ \ | '_ \
//| | | | (_) | | | |_| | | | | | | |
//|_| |_|\___/|_| \__|_| |_|___|_| |_|
// |_____|
#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define met_x(a) memset(a,0x3f,sizeof a)
#define mpy(a, b) memcopy(a,sizeof b,b)
#define int long long
#define ld long double
#define ull unsigned long long
#define fi first
#define se second
#define PII pair<int,int>
#define PDD pair<double,double>
#define PCI pair<char,int>
#define ALL(a) a.begin(),a.end()
#define rALL(a) a.begin(),a.end()
#define int128 __int128
#define endl '\n'
const int N = 200010;
const int M = 110;
const int MOD = 998244353;
const int EPS = 1e-8;
const int INF = 0x3f3f3f3f;
using namespace std;
int dp[N][2];
void solve() {
met_x(dp);
int n, a, b;
cin >> n >> a >> b;
string s;
cin >> s;
dp[0][0] = b;
for (int i = 0; i < n; i++) {
if (s[i] == '1')dp[i + 1][1] = dp[i][1] + a + b * 2;
else {
dp[i + 1][0] = min(dp[i][0] + a + b, dp[i][1] + a * 2 + b);
dp[i + 1][1] = min(dp[i][0] + a*2 + b * 2, dp[i][1] + a + b * 2);
}
}
cout << dp[n][0] << endl;
}
int32_t main() {
IOS;
int h_h = 1;
cin >> h_h;
while (h_h--)solve();
return 0;
}
D Number Of Permutations
正向求解问题非常难，所以我们可以使用容斥原理来求，先求第一个序列非递减的数量，再求第二序列非递减的数量，最后再求两个同时非递减的数量，最后总情况减去两个非递减的情况再将上两个同是非递减的数量，防止加多，这里要特别注意就是可能不存在两个数列有同事非递减的情况，我们按第一列排序，来判断第二列是否非递减即可，如果时非递减，俺么存在重合的，然后不是非递减，那么就不存在重合的
//
// Author : north_h
// File : D.cpp
// Time : 2023/7/24/15:10
// _ _ _
// _ __ ___ _ __| |_| |__ | |__
//| '_ \ / _ \| '__| __| '_ \ | '_ \
//| | | | (_) | | | |_| | | | | | | |
//|_| |_|\___/|_| \__|_| |_|___|_| |_|
// |_____|
#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define met_x(a) memset(a,0x3f,sizeof a)
#define mpy(a, b) memcopy(a,sizeof b,b)
#define int long long
#define ld long double
#define ull unsigned long long
#define fi first
#define se second
#define PII pair<int,int>
#define PDD pair<double,double>
#define PCI pair<char,int>
#define ALL(a) a.begin(),a.end()
#define rALL(a) a.begin(),a.end()
#define int128 __int128
#define endl '\n'
const int N = 300010;
const int M = 110;
const int MOD = 998244353;
const int EPS = 1e-8;
const int INF = 0x3f3f3f3f;
using namespace std;
int n;
int fact[N];
map<int,int> mp1,mp2;
map<PII,int> mp3;
vector<PII> a;
void solve() {
cin >> n;
for (int i = 1; i <= n; i++) {
int x, y;
cin >> x >> y;
mp1[x]++;
mp2[y]++;
mp3[{x, y}]++;
a.emplace_back(x, y);
}
fact[0] = 1;
for (int i = 1; i <= n; i++)fact[i] = fact[i - 1] * i % MOD;
int res1 = 1;
int res2 = 1;
int res3 = 1;
for (auto i: mp1)res1 = res1 * fact[i.se] % MOD;
for (auto i: mp2)res2 = res2 * fact[i.se] % MOD;
for (auto i: mp3)res3 = res3 * fact[i.se] % MOD;
sort(ALL(a));
bool ok = true;
for (int i = 1; i < a.size(); i++) if (a[i].se < a[i - 1].se)ok = false;
int ans = (fact[n] - res1 + MOD) % MOD;
ans = (ans - res2 + MOD) % MOD;
if (ok)ans = (ans + res3) % MOD;
cout << ans % MOD << endl;
}
int32_t main() {
IOS;
int h_h = 1;
//cin >> h_h;
while (h_h--)solve();
return 0;
}
E XOR Guessing
又是一道交互题，只有两次询问，要知道一个2^14-1的数，所以可以是使用位运算的知识，先求二进制前七位，再求二进制后七位，最后把得到的前七位和后七位拼在一起即可。
//
// Author : north_h
// File : E.cpp
// Time : 2023/7/24/15:10
// _ _ _
// _ __ ___ _ __| |_| |__ | |__
//| '_ \ / _ \| '__| __| '_ \ | '_ \
//| | | | (_) | | | |_| | | | | | | |
//|_| |_|\___/|_| \__|_| |_|___|_| |_|
// |_____|
#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(false),cin.tie(nullptr), cout.tie(nullptr);
#define met_0(a) memset(a,0,sizeof a)
#define met_1(a) memset(a,-1,sizeof a)
#define met_x(a) memset(a,0x3f,sizeof a)
#define mpy(a, b) memcopy(a,sizeof b,b)
#define ll long long
#define ld long double
#define ull unsigned long long
#define fi first
#define se second
#define PII pair<int,int>
#define PDD pair<double,double>
#define PCI pair<char,int>
#define ALL(a) a.begin(),a.end()
#define rALL(a) a.begin(),a.end()
#define int128 __int128
#define endl '\n'
const int N = 10010;
const int M = 110;
const int MOD = 998244353;
const int EPS = 1e-8;
const int INF = 0x3f3f3f3f;
using namespace std;
void solve() {
cout << "?" << ' ';
for (int i = 1; i <= 100; i++)cout << i << ' ';
cout.flush();
int n;
cin >> n;//高七位
cout << "?" << ' ';
for (int i = 1; i <= 100; i++)cout << (i << 7) << ' ';
cout.flush();
int m;
cin >> m;//低七位
int x = n & 16256;
x >>= 7, x <<= 7;
int y = m & 127;
y <<= 7, y >>= 7;
cout << "! ";
cout << (x | y) << endl;
}
int main() {
IOS;
int h_h = 1;
//cin >> h_h;
while (h_h--)solve();
return 0;
}
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