The "eight queens puzzle" is the problem of placing eight chess queens on an 8×8 chessboard so that no two queens threaten each other. Thus, a solution requires that no two queens share the same row, column, or diagonal. The eight queens puzzle is an example
of the more general N queens problem of placing N non-attacking queens on an N×N chessboard. (From Wikipedia - "Eight queens puzzle".)

Here you are NOT asked to solve the puzzles. Instead, you are supposed to judge whether or not a given configuration of the chessboard is a solution. To simplify the representation of a chessboard, let us assume that no two queens will be placed in the same
column. Then a configuration can be represented by a simple integer sequence (Q₁, Q₂, ..., Q_N), where Q_i is the row number of
the queen in the i-th column. For example, Figure 1 can be represented by (4, 6, 8, 2, 7, 1, 3, 5) and it is indeed a solution to the 8 queens puzzle; while Figure 2 can be represented by (4, 6, 7, 2, 8, 1, 9, 5, 3) and is NOT a 9 queens' solution.

Figure 1		Figure 2

Input Specification:

Each input file contains several test cases. The first line gives an integer K (1 < K <= 200). Then K lines follow, each gives a configuration in the format "N Q₁ Q₂ ... Q_N",
where 4 <= N <= 1000 and it is guaranteed that 1 <= Q_i <= N for all i=1, ..., N. The numbers are separated by spaces.

Output Specification:

For each configuration, if it is a solution to the N queens problem, print "YES" in a line; or "NO" if not.

Sample Input:

4

8 4 6 8 2 7 1 3 5

9 4 6 7 2 8 1 9 5 3

6 1 5 2 6 4 3

5 1 3 5 2 4

Sample Output:

YES

NO

NO

YES

#include <iostream>

#include <cstdio>

#include <string>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <queue>

#include <vector>

#include <set>

#include <stack>

#include <map>

#include <climits>

using namespace std;

#define LL long long

const int INF = 0x3f3f3f3f;

int a[10005];

int main()

{

    int T,n;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d",&n);

        for(int i=1; i<=n; i++)

            scanf("%d",&a[i]);

        int flag=0;

        for(int i=1; i<=n; i++)

            for(int j=1; j<i; j++)

            {

                if(a[i]==a[j]||fabs(a[i]-a[j])==fabs(j-i))

                {

                    flag=1;

                    break;

                }

            }

        printf("%s\n",flag==1?"NO":"YES");

    }

    return 0;

}