Codeforces A. Omkar and Completion (思维 / 构造 / 暴力) (Round #655 Div.2)

传送门

题意： 需构造一个长度为n的序列，满足其中的数字都不能超过1000，且对应任意1 <= z，y，z <= n，都有ax + ay != az。

思路： 简单的思维题，既然不能ax + ay != az，那么就全部输出1吧！

代码实现：

#include
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int  inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll   mod = 1e9 + 7;
const int  N = 2e5 + 5;

int t, n;

signed main()
{
    IOS;
    
    cin >> t;
    while(t --){
        cin >> n;
        while(n --)
            cout << 1 << " ";
        cout << endl;
    }

    return 0;
}