1487C - Minimum Ties

// Problem: 密码锁
// Contest: AcWing
// URL: https://www.acwing.com/problem/content/1353/
// Memory Limit: 64 MB
// Time Limit: 1000 ms
// 2022-04-17 23:03:33
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);
inline ll ksc(ll x,ll y,ll mod)
{
    ll ans = 0;
    while (y) {
        if (y & 1)
            ans = (ans + x) %mod;
        y >>= 1;
        x = (x + x) %mod;
    }
    return ans;
}
inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a;
        a = a * a;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
// inline ll inv (ll a) {
    // return qmi(a, mod - 2);
// }
int x[3], y[3];
int n;
bool cmp (int a, int b, int c, int xx[]) {
    if (!(abs(xx[0] - a) <= 2 || abs(xx[0] - a) >= n - 2))
        return 0;
    if (!(abs(xx[1] - b) <= 2 || abs(xx[1] - b) >= n - 2))
        return 0;
    if (!(abs(xx[2] - c) <= 2 || abs(xx[2] - c) >= n - 2))
        return 0;
    return 1;
}
void solve() {
//  int n;
    cin >> n;
    if (n % 2) {
        for (int i =1 ; i<= n; i ++) {
            for (int j = i + 1;j <= n; j ++)
                if (j - i <= n / 2)
                    cout << 1 <<" ";
                else 
                    cout << -1 << ' ';
        }
        puts("");
    }
    else {
        for (int i =1; i <= n; i ++) {
            for (int j = i + 1; j<= n; j ++) {
                if (j - i < n / 2) cout << 1 << " ";
                else if (j - i == n /2)
                    cout << 0 << ' ';
                else cout << "-1 ";
            }
        }
        puts("");
    }
    
}
int main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}

对于n为奇数，我们可以令一半赢一般输，因为每个只比一场，如果最终要相等，那必须每个都要赢一场，对于当前的i 对于n为偶数，由于n * (n - 1) * 3 / 2 mod n !=0,所以需要让它减去t，使得能够被n整除 (n * (n - 1) * 3 / 2 - n / 2) % n == 0总共要减去n / 2次，这可以通过每次j - i == n / 2的位置实现 需要找到总分数的这个性质，以及对于偶数需要找到关键问题所在，那就是不能被n整除C. Minimum Ties