传送门
A 雨
签到题，可惜我还多打了一个空格，不好说
/*
Author : north_h
File : B.cpp
Time : 2023/7/26/14:58
_ _ _
_ __ ___ _ __| |_| |__ | |__
| '_ \ / _ \| '__| __| '_ \ | '_ \
| | | | (_) | | | |_| | | | | | | |
|_| |_|\___/|_| \__|_| |_|___|_| |_|
|_____|
*/
#pragma GCC optimize(3)
#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 a, b, c, d;
cin >> a >> b >> c >>d;
int x;
cin >> x;
if (a < x)cout << x - a << ' ';
else cout << 0 << ' ';
if (b < x)cout << x - b << ' ';
else cout << 0 << ' ';
if (c < x)cout << x - c << ' ';
else cout << 0 << ' ';
if (d < x)cout << x - d << ' ';
else cout << 0 << ' ';
}
int main() {
IOS;
int h_h = 1;
//cin >> h_h;
while (h_h--)solve();
return 0;
}
B 吻
推公式，可能会报longlong，用int128来算，最后输出的时候转成longlong即可
/*
Author : north_h
File : A.cpp
Time : 2023/7/26/14:58
_ _ _
_ __ ___ _ __| |_| |__ | |__
| '_ \ / _ \| '__| __| '_ \ | '_ \
| | | | (_) | | | |_| | | | | | | |
|_| |_|\___/|_| \__|_| |_|___|_| |_|
|_____|
*/
#pragma GCC optimize(3)
#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 = 10010;
const int M = 110;
const int MOD = 998244353;
const int EPS = 1e-8;
const int INF = 0x3f3f3f3f;
using namespace std;
void solve() {
int n;
cin >> n;
int128 res = (int128) (1 + 2 * n - 1) * n / 2 % MOD;
cout << (int) res << endl;
}
int32_t main() {
IOS;
int h_h = 1;
//cin >> h_h;
while (h_h--)solve();
return 0;
}
C 失
按照题目意思模拟一下即可
/*
Author : north_h
File : C.cpp
Time : 2023/7/26/14:58
_ _ _
_ __ ___ _ __| |_| |__ | |__
| '_ \ / _ \| '__| __| '_ \ | '_ \
| | | | (_) | | | |_| | | | | | | |
|_| |_|\___/|_| \__|_| |_|___|_| |_|
|_____|
*/
#pragma GCC optimize(3)
#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;
bool cmp(pair<double,string> c,pair<double,string>d) {
if (c.fi != d.fi)return c.fi > d.fi;
else return c.se < d.se;
}
void solve() {
string s;
int n;
cin >> s >> n;
vector<string> a(n);
vector<pair<double, string>> ans;
for (auto &i: a)cin >> i;
for (auto i: a) {
int cnt = 0;
if (i.size() != s.size()) {
ans.push_back({0, i});
continue;
}
for (int j = 0; j < s.size(); j++) {
if (s[j] == i[j])cnt++;
}
int sum = s.size();
ans.push_back({cnt * 1.0 / sum, i});
}
sort(ALL(ans), cmp);
double now = ans[0].fi;
cout << ans[0].se << endl;
for (int i = 1; i < ans.size(); i++) {
if (ans[0].fi == now)cout << ans[i].se << endl;
}
}
int main() {
IOS;
int h_h = 1;
//cin >> h_h;
while (h_h--)solve();
return 0;
}
D 吹
最简单的动态规划，没写出来。菜是原罪，这里只有两个状态，要么前面变成1，要么不变，但是每个位置都要讨论变和不变的情况，每种情况都由前面变与不变的情况推出来
/*
Author : north_h
File : D.cpp
Time : 2023/7/26/14:58
_ _ _
_ __ ___ _ __| |_| |__ | |__
| '_ \ / _ \| '__| __| '_ \ | '_ \
| | | | (_) | | | |_| | | | | | | |
|_| |_|\___/|_| \__|_| |_|___|_| |_|
|_____|
*/
#pragma GCC optimize(3)
#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 = 100010;
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() {
int n;
cin >> n;
vector<int> a(n + 1);
for (int i = 1; i <= n; i++)cin >> a[i];
for (int i = 2; i <= n; i++) {
dp[i][0] = max(dp[i - 1][0] + abs(a[i] - a[i - 1]), dp[i - 1][1] + abs(a[i] - 1));
dp[i][1] = max(dp[i - 1][0] + a[i - 1] - 1, dp[i - 1][1]);
}
cout << max(dp[n][1], dp[n][0]) << endl;
}
int main() {
IOS;
int h_h = 1;
//cin >> h_h;
while (h_h--)solve();
return 0;
}
E 唤
这才是纯纯的贪心，首先456的方块看到要独自占一个，剩下的用1和2去补齐，3的方块全部用来构造6的方块，不够的那也用1和2去补齐，如果2填完了前面的还多出来，2还要单独去构成6的方块，不够的用1去补齐，最后1填完了前面的多出来，再用1去构造6的方块
/*
Author : north_h
File : E.cpp
Time : 2023/7/26/14:58
_ _ _
_ __ ___ _ __| |_| |__ | |__
| '_ \ / _ \| '__| __| '_ \ | '_ \
| | | | (_) | | | |_| | | | | | | |
|_| |_|\___/|_| \__|_| |_|___|_| |_|
|_____|
*/
#pragma GCC optimize(3)
#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 = 10010;
const int M = 110;
const int MOD = 998244353;
const int EPS = 1e-8;
const int INF = 0x3f3f3f3f;
using namespace std;
void solve() {
int s;
cin >> s;
int a1, a2, a3, a4, a5, a6;
cin >> a1 >> a2 >> a3 >> a4 >> a5 >> a6;
int sum = a6 + a5 + a4 + (a3 + 3) / 4;
int cnt2 = a4 * 5;
if (a3 % 4 == 1)cnt2 += 5;
else if (a3 % 4 == 2)cnt2 += 3;
else if (a3 % 4 == 3)cnt2++;
int cnt1 = 11 * a5;
if (a3 % 4 == 1)cnt1 += 7;
else if (a3 % 4 == 2)cnt1 += 6;
else if (a3 % 4 == 3)cnt1 += 5;
if (a2 > cnt2) {
a2 -= cnt2;
int cnt = (a2 + 8) / 9;
sum += cnt;
int y = (cnt * 9 - a2);
cnt1 += y * 4;
} else {
int cnt = cnt2 - a2;
cnt1 += cnt * 4;
}
a1 = max(0ll, a1 - cnt1);
sum += (a1 + 35) / 36;
if (sum <= s)cout << "Yes" << endl;
else cout << "No" << endl;
}
int32_t main() {
IOS;
int h_h = 1;
cin >> h_h;
while (h_h--)solve();
return 0;
}
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