Mendeleev 1417_牛客博客

$n = 4$ 时好方案是 $C_{4}^{2} + C_{4}^{4} = 7$ ，坏方案是 $C_{4}^{1} + C_{4}^{3} = 8\\$ $n = 5$ 时好方案是 $C_{5}^{2} + C_{5}^{4} = 15$ ，坏方案是 $C_{5}^{1} + C_{5}^{3} + C_{5}^{5} = 16 \\$ $n = 6$ 时好方案是 $C_{6}^{2} + C_{6}^{4} + C_{6}^{6}= 31$ ，坏方案是 $C_{6}^{1} + C_{6}^{3} + C_{6}^{5} = 32 \\$ $\\$ 显然的结论就是坏方案一定比好方案多1 $\\$

#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
# include<iostream>
# include<iomanip>
# include<algorithm>
# include<cmath>
# include<cstdio>
# include<set>
# include<stack>
# include<queue> 
# include<map>
# include<string>
# include<cstring> 
 
# define eps 1e-9
# define fi first
# define se second
# define ll long long
# define int ll
// cout<<fixed<<setprecision(n) 
using namespace std;
 
typedef unsigned long long ull;
typedef pair<int,int > PII; 
const int mod=1e9+7;
const int MAX=4e5+10;
const int Time=86400;
const int X=131;
const int inf=0x3f3f3f3f;
const double PI = 1e-4;
double pai = 3.14159265358979323846; 

int T,n,k,ans,m1;

void solve(){
     cin >> n;
     cout<<"-1\n";
}


signed main(){  
    std::ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> T;
    while(T--){
    	solve();
	}

    return 0;
}
/*1 2 3 4

1 2 3 4
1 2
1 3
1 4
2 3
2 4
3 4


1
2
3
4
1 2 3
1 2 4
1 3 4
2 3 4*/