Codeforces Round #839 (Div. 3)题解

A. A+B?

直接读入字符串然后把下标0和2的数字提取出来就行

// Problem: A. A+B?
// Contest: Codeforces - Codeforces Round #839 (Div. 3)
// URL: https://codeforces.com/contest/1772/problem/A
// Memory Limit: 512 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)// #pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
// #define int long long
#define endl '\n'
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define C2(n) (n * (n - 1) >> 1)
#define ll long long
#define ull unsigned long long
#define PII pair<int, int>
#define vint vector<int>
#define pb push_back
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define eqs 1e-6
// const int mod =
// const int N = void solve()
{string s;cin >> s;int a = s[0] - '0',b = s[2] - '0';cout << a + b<< endl;
}signed main()
{ios::sync_with_stdio(false);cin.tie(0);int t = 1;cin >> t;while(t--)solve();return 0;
}

B. Matrix Rotation

最大值和最小值处在对角线上时即为YES，或者像我这样列举所有情况

// Problem: B. Matrix Rotation
// Contest: Codeforces - Codeforces Round #839 (Div. 3)
// URL: https://codeforces.com/contest/1772/problem/B
// Memory Limit: 512 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)// #pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
// #define int long long
#define endl '\n'
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define C2(n) (n * (n - 1) >> 1)
#define ll long long
#define ull unsigned long long
#define PII pair<int, int>
#define vint vector<int>
#define pb push_back
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define eqs 1e-6
// const int mod =
// const int N = vector<int> get(vector<int> a)
{int k = min({a[0],a[1],a[2],a[3]});vint b = a;if(a[1] == k)b[0] = a[1],b[1] = a[3],b[2] = a[0],b[3] = a[2];else if(a[2] == k)b[0] = a[2],b[1] = a[0],b[2] = a[3],b[3] = a[1];else if(a[3] == k)b[0] = a[3],b[1] = a[2],b[2] = a[1],b[3] = a[0];return b;
}void solve()
{vint a(4);cin >> a[0] >> a[1] >> a[2] >> a[3];a = get(a);if(a[0] < a[1] && a[2] < a[3] && a[0] < a[2] && a[1] < a[3])puts("YES");elseputs("NO");
}signed main()
{ios::sync_with_stdio(false);cin.tie(0);int t = 1;cin >> t;while(t--)solve();return 0;
}

C. Different Differences

先从小到大标记：1、2、4、7……，不足的在从大到小标记

// Problem: C. Different Differences
// Contest: Codeforces - Codeforces Round #839 (Div. 3)
// URL: https://codeforces.com/contest/1772/problem/C
// Memory Limit: 512 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)// #pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
// #define int long long
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define C2(n) (n * (n - 1) >> 1)
#define ll long long
#define ull unsigned long long
#define PII pair<int, int>
#define vint vector<int>
#define pb push_back
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define eqs 1e-6
// const int mod =
// const int N =void solve()
{int n, m;cin >> n >> m;int res = 0;vint a(m + 1, 0),ans;for (int i = 1; i <= m; res++){if (res == n)break;a[i] = true;i += res + 1;}for (int i = m; i; --i){if (res == n)break;if (!a[i]){a[i] = true;res++;}}for(int i = 1;i <= m;++i){if(a[i])cout << i << ' ';}cout << endl;
}signed main()
{ios::sync_with_stdio(false);cin.tie(0);int t = 1;cin >> t;while (t--)solve();return 0;
}

D. Absolute Sorting

我是撒比，这题开始用二分答案来写，卡了一个半小时。正确解法是枚举相邻两位的差来确定取值范围即可

// Problem: D. Absolute Sorting
// Contest: Codeforces - Codeforces Round #839 (Div. 3)
// URL: https://codeforces.com/contest/1772/problem/D
// Memory Limit: 512 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)// #pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define C2(n) (n * (n - 1) >> 1)
#define ll long long
#define ull unsigned long long
#define PII pair<int, int>
#define vint vector<int>
#define pb push_back
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define eqs 1e-6
// const int mod =
// const int N =void solve()
{int n;cin >> n;vint a(n);for (int i = 0; i < n; ++i)cin >> a[i];int l = 0, r = inf;for (int i = 1, c; i < n; ++i){if (a[i - 1] < a[i]){c = (a[i - 1] + a[i]) / 2;r = min(r, c);}else if (a[i - 1] > a[i]){c = (a[i - 1] + a[i] + 1) / 2;l = max(l, c);}}if (l <= r)cout << l << endl;elsecout << -1 << endl;
}signed main()
{ios::sync_with_stdio(false);cin.tie(0);int t = 1;cin >> t;while (t--)solve();return 0;
}

E. Permutation Game

这题的题意有点含糊，其大意是，两个操作，将一个红色染成蓝色，或者将蓝色的位置之间任意排序。如：

此时可以选择3或者5将其染成蓝色，这里不做演示。也能对已经染成蓝色的进行重新排布，如：

当然也可以选择不操作

上述例子中，后手想要赢的话，只要将3、4、5都染成蓝色，那么他就可以将这三个数从新排布得到降序的数组；而先手想要赢的话，就要将所有的都染成蓝色，所以后手赢。

所以这题的重点在于染色上而不是排序上。我们可以将数分为三类，一类是只有先手想要染的（数量为a），第二类是只有后手想要染的（数量为b），第三类是两者都要染的（数量为c）。

所以当

a + c <= b时，先手胜（由于是先手，所以可以取等于）
b + c < a时，后手胜
平局

// Problem: E. Permutation Game
// Contest: Codeforces - Codeforces Round #839 (Div. 3)
// URL: https://codeforces.com/contest/1772/problem/E
// Memory Limit: 256 MB
// Time Limit: 4000 ms
//
// Powered by CP Editor (https://cpeditor.org)// #pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
// #define int long long
#define endl '\n'
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define C2(n) (n * (n - 1) >> 1)
#define ll long long
#define ull unsigned long long
#define PII pair<int, int>
#define vint vector<int>
#define pb push_back
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define eqs 1e-6
// const int mod =
// const int N = void solve()
{int n;cin >> n;int num[3]{0,0,0};for(int i = 1,j = n,c;i <= n;++i,--j){cin >> c;num[1] += (c != i);num[2] += (c != j);num[0] += (c != i && c != j);}if(num[1] <= num[2] - num[0])puts("First");else if(num[2] < num[1] - num[0])puts("Second");elseputs("Tie");
}signed main()
{ios::sync_with_stdio(false);cin.tie(0);int t = 1;cin >> t;while(t--)solve();return 0;
}

c;
num[1] += (c != i);
num[2] += (c != j);
num[0] += (c != i && c != j);
}
if(num[1] <= num[2] - num[0])
puts(“First”);
else if(num[2] < num[1] - num[0])
puts(“Second”);
else
puts(“Tie”);
}

signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);

int t = 1;
cin >> t;
while(t--)solve();return 0;

}

Codeforces Round #839 (Div. 3)题解相关推荐

Codeforces Round #514 (Div. 2)题解
Codeforces Round #514 (Div. 2)题解 A 喵,直接模拟. B 枚举所有盖章时的,合法的,左上角的位置.能盖的话就盖一下.最后check一下图案是否相等即可 C 一轮一轮的扔 ...
Codeforces Round #182 (Div. 1)题解【ABCD】
Codeforces Round #182 (Div. 1)题解 A题:Yaroslav and Sequence1 题意: 给你\(2*n+1\)个元素,你每次可以进行无数种操作,每次操作必须选择其 ...
【算法题解】Codeforces Round #817 (Div. 4)题解
文章目录 Codeforces Round #817 (Div. 4)题解 A. Spell Check B. Colourblindness C. Word Game D. Line E. Coun ...
Codeforces Round #747 (Div. 2)题解
Codeforces Round #747 (Div. 2)题解 (本博客将持续更新以后每场CF div2的题解,喜欢ACM.OI的小伙伴记得点个关注哟) 昨天夜晚刷网络流刷入迷了,渐渐就忘记了我还要 ...
Codeforces Round #839 (Div. 3)
Problem - G - Codeforces (1)题目大意一个人想提升下棋的rating,但是他只能一轮一轮来,若是他大于或者等于对战的那个人的rating,他的rating就会加1,那个对战 ...
Codeforces Round #789 (Div. 2)题解
Codeforces Round #789 (Div. 2)题解 A. Tokitsukaze and All Zero Sequence 原题链接算法标签贪心排序思路情况一:数组存在零 → ...
Codeforces Round #748 (Div. 3) 题解完整A~G
Codeforces Round #748 (Div. 3) 题解 A. Elections 题意已知竞选中三个候选人的当前得票数 a , b , c a,b,c a,b,c,现在可以增加任何一个人 ...
Codeforces Round #533 (Div. 2)题解
link orz olinr AK Codeforces Round #533 (Div. 2) 中文水平和英文水平都太渣..翻译不准确见谅 T1.给定n<=1000个整数,你需要钦定一个值t, ...
Codeforces Round #734 (Div. 3) 题解
Hello大家好,今天给大家带来的是 Codeforces Round #734 (Div. 3) 的全题目讲解. 本文链接:https://www.lanqiao.cn/questions/2040 ...