$A. Most Unstable Array$
$solution$
给定$n,m$，要构造一个长度为$n$的数组，和为$m$，求$\sum_{i=1}^{n-1}\left|a_{i}-a_{i+1}\right|$最大
长度为$1$是$0$，为$2$是$m$，其他是$2\times m$
$code$
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define per(i,a,n) for(int i=n-1;i>=a;i--)
#define fi first
#define se second
#define ll long long
#define pb push_back
typedef pair<long long,long long> pll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<long long> vll;
typedef double db;
const ll mod=1e9+7;
ll powmod(ll a,ll b,ll p){ll res=1;a%=p;while(b){if(b&1) res=res*a%p;a=a*a%p;b>>=1;}return res;}
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
int _;
int main(){
for(scanf("%d",&_);_;_--){
int n,k;
scanf("%d%d",&n,&k);
if(n==1) {puts("0");}
else if(n==2) {printf("%d\n",k);}
else{
printf("%d\n",k*2);
}
}
}
$B. Two Arrays And Swaps$
$solution$
给定两个长$n,1<=n<=30$的数组$a,b$，给定一个$k$，可以将$a$中$k$个替换成$b$中元素，求得最大的和
排序后双指针两个分别从最大最小判断即可
$code$
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define per(i,a,n) for(int i=n-1;i>=a;i--)
#define fi first
#define se second
#define ll long long
#define pb push_back
typedef pair<long long,long long> pll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<long long> vll;
typedef double db;
const ll mod=1e9+7;
ll powmod(ll a,ll b,ll p){ll res=1;a%=p;while(b){if(b&1) res=res*a%p;a=a*a%p;b>>=1;}return res;}
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
int _;
const int N=50;
int a[N],b[N];
int main(){
for(scanf("%d",&_);_;_--){
int n,k;
cin>>n>>k;
rep(i,0,n) cin>>a[i];
rep(i,0,n) cin>>b[i];
sort(a,a+n);
sort(b,b+n);
int ans=0;
for(int i=0,j=n-1;i<n;i++,j--) {
if(a[i]<b[j] && k){ans+=b[j];k--;}
else ans+=a[i];
}
printf("%d\n",ans);
}
}
$C. Board Moves$
$solution$
$n \times n, n$(是奇数) 的矩阵，初始每个格子都有$1$个东西，八相邻方向移动算一步，将所有的移动到一个格子要多少步
因为是奇数所以有正中心，从中间向外扩计算即可
$code$
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define per(i,a,n) for(int i=n-1;i>=a;i--)
#define fi first
#define se second
#define ll long long
#define pb push_back
typedef pair<long long,long long> pll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<long long> vll;
typedef double db;
const ll mod=1e9+7;
ll powmod(ll a,ll b,ll p){ll res=1;a%=p;while(b){if(b&1) res=res*a%p;a=a*a%p;b>>=1;}return res;}
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
int _;
int main(){
for(scanf("%d",&_);_;_--){
ll n;
cin>>n;
if(n==1){puts("0");continue;}
n/=2;
ll ans=0;
for(ll i=1;i<=n;i++){
ans+=i*8*i;
}
cout<<ans<<endl;
}
}
$D. Constructing the Array$
$solution$
给定长度为$n,1<=n<=2 \times 10^{5}$的全0数组，选取长度最大的，长度相等的选取最靠做的连续子区间$[l,r]$
一个操作变量初始为1，如果区间长度为偶数，就将数组$(l+r-1)/2$位置上值改成这个变量
奇数就在$(l+r)/2$位置上记录，因为区间没有重叠，优先队列，重载比较函数，长度大的优先，相等左边小的优先
$code$
#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define per(i,a,n) for(int i=n-1;i>=a;i--)
#define fi first
#define se second
#define ll long long
#define pb push_back
typedef pair<long long,long long> pll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<long long> vll;
typedef double db;
const ll mod=1e9+7;
ll powmod(ll a,ll b,ll p){ll res=1;a%=p;while(b){if(b&1) res=res*a%p;a=a*a%p;b>>=1;}return res;}
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
int _;
const int N=2e5+10;
struct node{
int x,y;
bool operator <(const node &b) const{
if(x == b.x)
return y > b.y;
return x<b.x;
}
};
int main(){
for(scanf("%d",&_);_;_--){
int n;
cin>>n;
vi res(n+1);
priority_queue<node> seg;
int l=1,r=n;
int now=1;
seg.push({r-l+1,l});
while(seg.size()){
auto t=seg.top();
seg.pop();
if(t.x!=0){
r=t.y+t.x-1;
l=t.y;
int mid;
if((r-l+1)%2==0) {
res[(l+r-1)/2]=now;
mid=(l+r-1)/2;
}
else {
res[(l+r)/2]=now;
mid=(l+r)/2;
}
now++;
if(mid-1-l+1 != 0) seg.push({mid-1-l+1,l});
if(r-(mid+1)+1 !=0) seg.push({r-(mid+1)+1,mid+1});
}
}
for(int i=1;i<=n;i++) cout<<res[i]<<' ';
puts("");
}
}
$E. K-periodic Garland$
$solution$
$code$
$F. Decreasing Heights$