@[TOC](2020牛客暑期多校训练营（第三场）D——Points Construction Problem)
题目描述
输入描述
The first line contains one integer t (1≤t≤10 3) — the number of test cases.
The only line of each test case contains two integers n and m (1≤n≤50,1≤m≤200).
输出描述
For each test, if there exists at least one configuration to choose n points to satisfy the conditions given by statement, you should print n+1 line for this test. The first line contains one string “Yes”. And the following n lines contain the coordinator of these n points which is colored as black. If there are no solution, please print one line containing only one string “No”.
输入
6
5 20
1 2
1 3
1 4
1 5
3 8
输出
Yes
1 1
2 2
3 3
4 4
5 5
No
No
Yes
1 1
No
Yes
1 1
1 2
2 1
说明
In the first test and fourth test, each black point in the sample output is adjacent to exactly 4 white points.
In the sixth test, the second and third black points in the sample output are both adjacent to 3 white points and the forst black point is adjacent to 2 white points.
题目大意
假设您有一个具有笛卡尔坐标系的无限二维平面。 最初，所有整数点（xy坐标都为整数的点）都绘制为白色。 给出两个整数n和m。 请精确地将n个整数点涂成黑色，使得刚好有m个点对满足以下条件：
1.这两点用不同的颜色着色。
2.这两点相邻。 我们称两个整数点（x1，y1）和（x2，y2）相邻，当且仅当∣x1-x2∣ + ∣y1 −y2∣ = 1。 （| v |表示v的绝对值。）
3.所有黑点的x和y坐标在[-10 ^ 9，10 ^ 9] 范围内。
对于每项测试，如果至少存在一个配置来选择n个点来满足语句给出的条件，则应打印此测试的n+1行。第一行包含一个字符串“Yes”。并且下面的n行包含这n个黑色点的坐标。如果没有解决方案，请打印一行，该行仅包含一个字符串“No”。
题解
每个黑格子的贡献只能是4,2,0所以答案只能是偶数。我们可以类比构造来实现本题。
*注：本题有特殊评判机制！
AC代码
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll T,n,m,i,j,k,l,o,p,a[10010][10010];
int main()
{
//ios::sync_with_stdio(false);
for(scanf("%lld",&T);T--;)
{
scanf("%lld%lld",&n,&m);
if(m&1||m>4*n||n*16>m*m) puts("No");
else
{
puts("Yes");
if (m>2*n+2)
{
o=(m-(2*n+2))/2;p=n-o;
for (i=1;i<=p;i++) printf("1 %lld\n",i);
for (i=1;i<=o;i++) printf("4 %lld\n",i*2);
}
else
{
o=m/4,p=m/2-o;
for(i=1;i<=o;i++) printf("%lld 1\n",i);
for(j=2;j<=p;j++) printf("1 %lld\n",j);
l=n-(o+p-1);
for(i=2;i<=o&&l>0;i++)
for(j=2;j<=p&&l>0;j++,l--)
printf("%lld %lld\n",i,j);
}
}
}
}