Codeforces Round 884 (Div. 1 + Div. 2)（视频讲解A--D）

@[TOC](Codeforces Round 884 (Div. 1 + Div. 2)（视频讲解A–D）)

视频链接：Codeforces Round 884 (Div. 1 + Div. 2)（视频讲解A–D）

A Subtraction Game

1、 板书：

2、代码

#include<bits/stdc++.h>
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N = 1e5 + 10;void solve()
{int a, b;cin >> a >> b;cout << a + b << endl;	
}signed main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t;cin >> t;while(t--)solve();
}

B Permutations & Primes

1、板书

2、代码

#include<bits/stdc++.h>
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N = 1e5 + 10;void solve()
{int n;cin >> n;vector<int>a(n);if(n <= 2){for(int i = 1; i <= n; i ++){	cout << i << " ";}cout << endl;return;}a[(n - 1) / 2] = 1;a[0] = 2;a[n - 1] = 3;int tmp = 4;for(int i = 0; i < n; i ++)if(a[i] == 0)a[i] = tmp ++;for(int i = 0; i < n; i ++){cout << a[i] << " ";}cout << endl;
}signed main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t;cin >> t;while(t--)solve();
}

C Particles

1、板书

2、代码

#include<bits/stdc++.h>
#define endl '\n'
#define INF 0x3f3f3f3f
#define int long long
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N = 1e5 + 10;void solve()
{int n;cin >> n;vector<int>a, b;for(int i = 0; i < n; i ++){int x;cin >> x;if(i % 2)a.push_back(x);elseb.push_back(x);}if(n == 1){cout << b[0] << endl;return;}sort(a.begin(), a.end());sort(b.begin(), b.end());int ans1 = 0, ans2 = 0;if(a[a.size() - 1] < 0)ans1 = a[a.size() - 1];else{for(int i = a.size() - 1; i >=0; i --){if(a[i] > 0)ans1 += a[i];elsebreak;}}if(b[b.size() - 1] < 0)ans2 = b[b.size() - 1];else{for(int i = b.size() - 1; i >=0; i --){if(b[i] > 0)ans2 += b[i];elsebreak;}}cout << max(ans1, ans2) << endl;}signed main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t;cin >> t;while(t--)solve();
}

D Row Major

1、板书

2、代码

#include<bits/stdc++.h>
#define endl '\n'
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N = 1e5 + 10;void solve()
{int n;cin >> n;int x;for(int i = 1; i <= 26; i ++){if(n % i){x = i;break;}}for(int i = 0; i < n; i ++){char c = 'a' + (i % x);cout << c;}cout << endl;
}signed main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t;cin >> t;while(t--)solve();
}