2019牛客多校第七场B题

本题问你能否将一个已知系数的多项式因式分解，我们只要知道一个性质就ok；
一个多项式不可分解，当且仅当它次数为1或者次数为2但判别式小于零。由此可知：

#include<cstdio>
#include <algorithm>
#include <iostream>
#include <memory.h>
#include <queue>
#define rep(i, n) for(int i=0;i<n;i++)
#define per(i, n) for(int i=n;i>=1;i--)
#define pb(x) push_back(x)
#define clint(x, n) memset(x,n,sizeof(x))
typedef long long  ll;
const int maxn = 25;
const ll inf = 99999999;
using namespace std;
int dirx[] = {1, 0, -1, 0, 0, 0};
int diry[] = {0, 1, 0, -1, 0, 0};
int dirz[] = {0, 0, 0, 0, 1, -1};
int px, py, pz;
int m, n, T, t;
char ma[30][30][30];
int vis[30][30][30];
int ans;
int ex, ey, ez;
char ss[10];
string str;
int a[1010];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--) {
        int n;
        while (~scanf("%d", &n)) {
            for (int i = n; i >= 0; i--) scanf("%d", &a[i]);
            if (n < 2) cout << "Yes" << endl;
            else if (n == 2) {
                if ((a[1] * a[1]) >= 4 * a[2] * a[0]) cout << "No" << endl;
                else cout << "Yes" << endl;
            } else {
                cout << "No" << endl;
            }
        }
    }
    return 0;
}