1.题目描写叙述：点击打开链接

2.解题思路：本题是四色定理的模板题。只是有几种情况要提前特判一下：n==1直接输出，1<n<5时候无解，n==6时候套用模板会出现同样的块。因此要特判一下。其它情况都能直接利用模板构造出来。

3.代码：

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<algorithm>
#include<cassert>
#include<string>
#include<sstream>
#include<set>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<cctype>
#include<functional>
using namespace std;

#define me(s) memset(s,0,sizeof(s))
#define pb push_back
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair <int, int> P;



const int N = 117;
char mm[N][N];
char col[2];
int n;

void solve()  //四色定理的模板
{
	memset(mm, '.', sizeof(mm));
	for (int i = 0; i<n; i++)
	for (int j = 0; j<n; j++)
		mm[i][j] = 'B';
	for (int i = 0; i<n; i++)
		mm[0][i] = 'Y';
	for (int i = 0; i<(n - 1) / 2; i++)
	{
		char c = col[i % 2];
		for (int j = i + 1; j<n; j++)
			mm[j][i] = c;
		for (int j = 1; j <= i + 1; j++)
			mm[j][i + 1] = c;
	}
	for (int i = (n - 1) / 2; i<n - 2; i++)
	{
		char c = col[i % 2];
		for (int j = i + 2; j<n; j++)
			mm[j][i] = c;
		mm[i + 2][i + 1] = c;
		for (int j = 2; j <= i + 2; j++)
			mm[j][i + 2] = c;
	}
	for (int i = 0; i<n; i++)
	{
		for (int j = 0; j<n; j++)
			printf("%c", mm[i][j]);
		printf("\n");
	}
}

int main()
{
	col[0] = 'G';
	col[1] = 'R';
	int T;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d", &n);
		if (n == 1)printf("Y\n");
		else if (n <= 4)
			printf("No solution!\n");
		else if (n == 6)
		{
			printf("YYYYYY\n");
			printf("GGRGRR\n");
			printf("GRRGRB\n");
			printf("GRGGRB\n");
			printf("GRGRRB\n");
			printf("GRGBBB\n");
		}
		else solve();  //n==5和n>6时候能够直接使用模板构造
	}
}