code forces round 1008 (div. 2) (C-D)

Update E later

Problem - C - Codeforces

Ideas

It can be found that one of the conditions in the question is actuallyThe sum of odd terms equals even terms

So we can sort 2n numbers, take the previous one to do odd numbers, and sum it\(pre\), do even numbers afterwards, and\(suf\),So\(a_1\)that is\(suf-pre\)Meet this condition

But just this may not be satisfiedDifferent twosThe condition of the previous\(a_1\)equal to the given\(2n\)The number of
The method of handling is very simple, let's get it\(n-1\)Add the number\(suf-pre\)Do odd terms,\([n,n*2-1]\)The number is an even number, and the last number is\(a_1\)Just

For example, the last example

Code

void solve(){
    int n,k;cin>>n;
    vector<int>a(n*2+1);
    rep(i,1,n*2) cin>>a[i];
    sort(()+1,());
    int pre=0,suf=0;
    rep(i,1,n) pre+=a[i];
    per(i,n*2,n+1) suf+=a[i];
    pre-=a[n];
    suf+=a[n];
    int tmp=suf-pre;
    cout<<a[n*2]<<" ";
    vector<int>b(n*2+1);
    int j=1,s=n,f=0;
    rep(i,1,n*2){
        if(i&1){
            if(f==0) b[i]=tmp,f=1;
            else b[i]=a[j],j++;
        } 
        else b[i]=a[s],s++;
    }
    _db(b,n*2);
}

Problem - D - Codeforces

Ideas

Greedy, forAll are added or multiplied, which side is placed in the current step, the contribution to the answer is the same, but it may lead to the lack of advantages in the future.
All the contributions here are betterTemporary storageGet up, wait until other situations are needed before making a decision on the better side of the current step

Code

struct node{
    char c1;
    int a;
    char c2;
    int b;
};
void solve(){
    int n;cin>>n;
    vector<node>p(n+1);
    rep(i,1,n){
        char c1,c2;
        int a,b;
        cin>>c1>>a>>c2>>b;
        p[i]={c1,a,c2,b};
    }
    int l=1,r=1,now=0;
    rep(i,1,n){
        auto rt=p[i];
        int f=0;
        if(rt.c1=='x'&&rt.c2=='x'){
            if(>) l+=now;
            else if(<) r+=now;
            else{
                f=1;
                now=(l+r)*(-1)+now*;
            }
        }
        else if(rt.c1=='x') l+=now;
        else if(rt.c2=='x') r+=now;
        else{
            f=1;
            now+=+;
        }
        if(f) continue;
        now=0;
        if(rt.c1=='x') now+=(-1)*l;
        else now+=;
        if(rt.c2=='x') now+=(-1)*r;
        else now+=;
        // cout<<l<<" "<<r<<" "<<now<<"\n";
    }
    int ans=l+r+now;
    cout<<ans<<"\n";
}