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_pairtypedef 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

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