AtCoder Grand Contest 025 Problem D
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AtCoder Grand Contest 025 Problem D##
Time Limit: 2 Sec
Memory Limit: 1024 MBDescription###
Takahashi is doing a research on sets of points in a plane. Takahashi thinks a set \(S\) of points in a coordinate plane is a good set when \(S\) satisfies both of the following conditions:
• The distance between any two points in \(S\) is not \(\sqrt{D1}\).
• The distance between any two points in \(S\) is not \(\sqrt{D2}\) .
Here, \(D1\) and \(D2\) are positive integer constants that Takahashi specified.
Let \(X\) be a set of points \((i,j)\) on a coordinate plane where \(i\) and \(j\) are integers and satisfy \(0≤i,j<2N\).
Takahashi has proved that, for any choice of \(D1\) and \(D2\), there exists a way to choose \(N^2\) points from \(X\) so that the chosen points form a good set. However, he does not know the specific way to choose such points to form a good set. Find a subset of \(X\) whose size is \(N^2\) that forms a good set.
Input###
• \(1≤N≤300\)
• \(1≤D1≤2×105\)
• \(1≤D2≤2×105\)
• All values in the input are integers.
Input is given from Standard Input in the following format:
\(N\) \(D_1\) \(D_2\)
Output###
Print N2 distinct points that satisfy the condition in the following format:
\(x_1 y_1\)
\(x_2 y_2\)
:
\(x_{N^2} y_{N^2}\)
Here, (xi,yi) represents the i-th chosen point. \(0≤xi,yi<2N\) must hold, and they must be integers. The chosen points may be printed in any order. In case there are multiple possible solutions, you can output any.
Sample Input 1###
2 1 2
Sample Output 1###
0 0
0 2
2 0
2 2
Sample Input 2###
3 1 5
Sample Output 2###
0 0
0 2
0 4
1 1
1 3
1 5
2 0
2 2
2 4
题目地址： AtCoder Grand Contest 025 Problem D
题目大意：
输⼊ \(n, d_1, d_2\)，
你要找到 \(n^2\) 个整点 \(x, y\) 满⾜ \(0 ≤ x, y < 2n\)。 并且找到的任意两个点距离，既不是 \(\sqrt{d1}，也不是 \sqrt{d2}\)。
题解：
这是个分析题⽬。 **
简单来说，所有距离为 \(\sqrt{d_1}\) 的点连边，可以得到⼀个⼆分图。\(d_2\) 同理。 **
这样可以把所有 \(4n^2\) 个点四分，⼀定有⼀块满⾜条件。 **
如果 d mod 2 = 1，如果 \(a^2 + b^2 = d\)，a 和 b ⼀定⼀奇⼀偶，按国际象棋⿊⽩染⾊即可。 **
如果 d mod 4 = 2，如果 \(a^2 + b^2 = d\)，a 和 b ⼀定都是奇数，⼀⾏⿊⾊，⼀⾏⽩⾊即可。 **
如果 d mod 4 = 0，把 2 × 2 的区域看成⼀个⼤格⼦，如此类推，对 d/4 进⾏如上考虑即可。**
AC代码
#include <cstdio>
using namespace std;
int n,d1,d2,s;
int f[620][620];
void work(int d){
int p=0;
while(d%4==0){
d/=4;
p++;
}
if(d&1){
for(int i=0;i<2*n;i++)
for (int j=0;j<2*n;j++)
if(((i>>p)+(j>>p))&1)
f[i][j]=1;
}else{
for(int i=0;i<2*n;i++)
for(int j=0;j<2*n;j++)
if((i>>p)&1)
f[i][j]=1;
}
}
int main(){
scanf("%d%d%d",&n,&d1,&d2);
work(d1);
work(d2);
for(int i=0;i<2*n;i++){
for(int j=0;j<2*n;j++){
if(s<n*n && !f[i][j]){
printf("%d %d\n",i,j);
s++;
}
}
}
return 0;
}
