Codeforces1487 C. Minimum Ties (mathematics, construction)

Title:

solution:

$$\begin{gathered} Setxtowinbenefittimesthe\; number,yisflatBureautimesthe\; number,Duetothatthere\; areteamsdividednumberwiththe\; same, \\ Thatithas:x+Y=\frac{n\ast (n-1)}{2},3x+2y=kn, \\ Solutionshave\; to:x=n\ast (k-n+1),Y=n\ast (\frac{3n-3}{2}-k), \\ Whennisan\; oddnumberwhen,Letk=\frac{3n-3}{2},ThistimeflatBureautimesthe\; number\; ofmostsmall, \\ That\; is,eachbranchteamWuShengLee\frac{n-1}{2}Times,PingBureau0times. \\ Whennisan\; evennumber\; ofthe\; time,Letk=\frac{3n-4}{2},ThistimeflatBureautimesthe\; number\; ofmostsmall, \\ That\; is,eachbranchteamWuShengLee\frac{n-2}{2}Times,PingBoard1times. \end{gathered}$$

code：

#include <bits/stdc++.h>
#define int long long
using namespace std;
int n;
void solve(){
    
    
    cin>>n;
    if(n%2==1){
    
    
        int win=(n-1)/2;
        for(int i=1;i<=n;i++){
    
    
            int cnt=0;
            for(int j=i+1;j<=n;j++){
    
    
                cnt++;
                if(cnt<=win)cout<<1<<' ';
                else cout<<-1<<' ';
            }
        }
    }else{
    
    
        int win=(n-2)/2;
        for(int i=1;i<=n;i++){
    
    
            int cnt=0;
            for(int j=i+1;j<=n;j++){
    
    
                cnt++;
                if(cnt<=win)cout<<1<<' ';
                else if(cnt==win+1)cout<<0<<' ';
                else cout<<-1<<' ';
            }
        }
    }
    cout<<endl;
}
signed main(){
    
    
    ios::sync_with_stdio(0);
    int T;cin>>T;
    while(T--){
    
    
        solve();
    }
    return 0;
}

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