由于最近这三天的数学建模，让我这个精力本来就不多的AI手更加力竭了，没注意到昨晚的cf，所以今天来补题了。
比赛连接：比赛传送门

A题：

You are doing a research paper on the famous Collatz Conjecture. In your experiment, you start off with an integer $x$, and you do the following procedure $k$ times:

• If $x$ is even, divide $x$ by $2$.
• Otherwise, set $x$ to $3\cdot x+1$.

For example, starting off with $21$ and doing the procedure $5$ times, you get $21\to 64\to 32\to 16\to 8\to 4$.

After all $k$ iterations, you are left with the final value of $x$. Unfortunately, you forgot the initial value. Please output any possible initial value of $x$.
Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 400$). The description of the test cases follows.

The first line of each test case contains two integers $k$ and $x$ ($1\le k,x\le 20$).
Output

For each test case, print any possible initial value on a new line. It can be shown that the answer always exists.
Note

In the first test case, since $1$ is odd, performing the procedure $k=1$ times results in $1\cdot 3+1=4$, so $1$ is a valid output.

In the second test case, since $10$ is even, performing the procedure $k=1$ times results in $\frac{10}{2}=5$, so $10$ is a valid output.

The third test case is explained in the statement.

通过观察不难发现，两种操作都是变为了偶数，所以一直在原来的基础上乘2即可（永远都只用操作一即可）。

// Problem: A. Collatz Conjecture
// Contest: Codeforces - Codeforces Round 1047 (Div. 3)
// URL: https://codeforces.com/contest/2137/problem/A
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>
using namespace std;
#define int long long 
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define pii pair<int,int>
#define fi first
#define se second
#define YES cout<<"YES"<<endl;
#define NO cout<<"NO"<<endl;
#define lbt(x) ((x) & (-x)) const int INF = 0x3f3f3f3f;
const int inf = 1e18; void solve()
{int n,k;cin>>k>>n;int an = n;for(int i=1;i<=k;i++) an *= 2LL;cout<<an<<endl;
//	cout<<fixed<<setprecision(x)<< ; 
}signed main()// Don't forget pre_handle!
{IOSint T=1;cin>>T;while(T--) solve(); return 0;
} 

B题

You are given a permutation${}^{\text{∗}}$ $p$ of size $n$.

Your task is to find a permutation $q$ of size $n$ such that $\mathrm{GCD}$${}^{\text{†}}$$(p_i+q_i,p_{i+1}+q_{i+1})\ge 3$ for all KaTeX parse error: Expected 'EOF', got '&' at position 9: 1 \leq i&̲lt;n. In other words, the greatest common divisor of the sum of any two adjacent positions should be at least $3$.

It can be shown that this is always possible.

${}^{\text{∗}}$A permutation of length $m$ is an array consisting of $m$ distinct integers from $1$ to $m$ in arbitrary order. For example, $[2,3,1,5,4]$ is a permutation, but $[1,2,2]$ is not a permutation ($2$ appears twice in the array), and $[1,3,4]$ is also not a permutation ($m=3$ but there is $4$ in the array).

${}^{\text{†}}$$\gcd (x,y)$ denotes the greatest common divisor (GCD) of integers $x$ and $y$.
Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($2\le n\le 2\cdot 10^5$).

The second line contains $n$ integers $p_1,p_2,\dots ,p_n$ ($1\le p_i\le n$).

It is guaranteed that the given array forms a permutation.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2\cdot 10^5$.
Output

For each test case, output the permutation $q$ on a new line. If there are multiple possible answers, you may output any.
Note

In the first test case, $\mathrm{GCD}(1+2,3+3)=3\ge 3$ and $\mathrm{GCD}(3+3,2+1)=3\ge 3$, so the output is correct.

这道题也很好想，最大公约数，又因为是排列，所以我们肯定不难想到用最大的和去和每个元素进行配对，这样所有的两个位置上两个排列中所对应的数相加的和都相同了，那么这个时候就是满足条件的排列了。

// Problem: B. Fun Permutation
// Contest: Codeforces - Codeforces Round 1047 (Div. 3)
// URL: https://codeforces.com/contest/2137/problem/B
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>
using namespace std;
#define int long long 
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define pii pair<int,int>
#define fi first
#define se second
#define YES cout<<"YES"<<endl;
#define NO cout<<"NO"<<endl;
#define lbt(x) ((x) & (-x)) const int INF = 0x3f3f3f3f;
const int inf = 1e18; void solve()
{int n;cin>>n;vector<int> a(n+1);for(int i=1;i<=n;i++) cin>>a[i];int sum = n + 1;for(int i=1;i<=n;i++) cout<<sum - a[i]<<' ';cout<<endl;
//	cout<<fixed<<setprecision(x)<< ; 
}signed main()// Don't forget pre_handle!
{IOSint T=1;cin>>T;while(T--) solve(); return 0;
} 

C题

You are given two integers $a$ and $b$. You are to perform the following procedure:

First, you choose an integer $k$ such that $b$ is divisible by $k$. Then, you simultaneously multiply $a$ by $k$ and divide $b$ by $k$.

Find the greatest possible even value of $a+b$. If it is impossible to make $a+b$ even, output $-1$ instead.
Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $a$ and $b$ ($1\le a,b\le a\cdot b\le 10^{18})$.
Output

For each test case, output the maximum even value of $a+b$ on a new line.
Note

In the first test case, it can be shown it is impossible for $a+b$ to be even.

In the second test case, the optimal $k$ is $2$. The sum is $2+4=6$.

分情况讨论即可

// Problem: C. Maximum Even Sum
// Contest: Codeforces - Codeforces Round 1047 (Div. 3)
// URL: https://codeforces.com/contest/2137/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>
using namespace std;
#define int long long 
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define pii pair<int,int>
#define fi first
#define se second
#define YES cout<<"YES"<<endl;
#define NO cout<<"NO"<<endl;
#define lbt(x) ((x) & (-x)) const int INF = 0x3f3f3f3f;
const int inf = 1e18; void solve()
{int a,b;cin>>a>>b;if((a * b) & 1) {cout<<a * b + 1<<endl;return ;}int ans = -1;if(b & 1){cout<<ans<<endl;return ;}if((a * (b / 2LL) + 2LL) & 1){cout<<ans<<endl;return ;}// if(a & 1LL)// {cout<<a * (b / 2LL) + 2LL<<endl;return ;// }// if(a % 2 == 0)// {// cout<<a * b// }
//	cout<<fixed<<setprecision(x)<< ; 
}signed main()// Don't forget pre_handle!
{IOSint T=1;cin>>T;while(T--) solve(); return 0;
} 

D题

Given an array $a$, let $f(x)$ be the number of occurrences of $x$ in the array $a$. For example, when $a=[1,2,3,1,4]$, then $f(1)=2$ and $f(3)=1$.

You have an array $b$ of size $n$. Please determine if there is an array $a$ of size $n$ such that $f(a_i)=b_i$ for all $1\le i\le n$. If there is one, construct it.
Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains an integer $n$ ($1\le n\le 2\cdot 10^5$).

The second line contains $n$ integers $b_1,b_2,\dots ,b_n$ ($1\le b_i\le n$).

It is guaranteed that the sum of $n$ over all test cases does not exceed $2\cdot 10^5$.
Output

For each test case, output $-1$ if there is no valid array $a$.

Otherwise, print the array $a$ of $n$ integers on a new line. The elements should satisfy $1\le a_i\le n$.
Note

In the first test case, it can be shown that no array matches the requirement.

In the second test case, $4$, $5$, $6$ appear $1,2,3$ times respectively. Thus, the output array is correct as $f(4),f(5),f(5),f(6),f(6),f(6)$ are $1,2,2,3,3,3$ respectively.

这道题用map嵌套pair可以省去很多步骤，思路也比较清楚

// Problem: D. Replace with Occurrences
// Contest: Codeforces - Codeforces Round 1047 (Div. 3)
// URL: https://codeforces.com/contest/2137/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>
using namespace std;
#define int long long 
#define endl '\n'
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define pii pair<int,int>
#define fi first
#define se second
#define YES cout<<"YES"<<endl;
#define NO cout<<"NO"<<endl;
#define lbt(x) ((x) & (-x)) const int INF = 0x3f3f3f3f;
const int inf = 1e18; void solve()
{int n;cin>>n;vector<int> a(n+1);for(int i=1;i<=n;i++) cin>>a[i];map<int,pii> mp;for(int i=1;i<=n;i++) mp[a[i]].fi ++,mp[a[i]].se = 0;int num = 0;for(auto &i : mp) num += i.se.fi;if(num != n){cout<<"-1"<<endl;return ;}for(auto &i : mp){int t1 = i.fi,t2 = i.se.fi;if(t2 % t1){cout<<"-1"<<endl;return ;}}int index = 0LL;map<int,int> cnt;map<int,bool> v;for(int i=1;i<=n;i++){if(!v[a[i]]) mp[a[i]].se = ++index;if(cnt[a[i]] == a[i]){cnt[a[i]] = 0;mp[a[i]].se = ++index;}cnt[a[i]] ++;v[a[i]] = true;cout<<mp[a[i]].se<<' ';}// for(int i=1;i<=n;i++) cout<<mp[a[i]].se<<' ';cout<<endl;
//	cout<<fixed<<setprecision(x)<< ; 
}signed main()// Don't forget pre_handle!
{IOSint T=1;cin>>T;while(T--) solve(); return 0;
}