Atcoder #Contest 025-D. Choosing Points

題意

找到\(\ n^2\)個點，滿足0 \(\leq\) x,y < 2n且兩點間距離d \(\not\) = \(\sqrt{x1}\) 或 \(\sqrt{x2}\)

題解

設 \(\ a^2\) + \(\ b^2\) = d

若\(\ d\equiv0\)\(\pmod{2}\),a和b必定一奇一偶,按國際象棋染色即可

若\(\ d\equiv1\)\(\pmod{2}\),a和b必定均為奇數,一行白,一行黑染色即可

若\(\ d\equiv0\)\(\pmod{4}\),将2*2的區域看成一個大格子,d疊代為d/4 進行如上考慮即可

調試記錄

% 的優先級高于 + 要加括号

#include <cstdio>

using namespace std;

int n, d1, d2, a[605][605];

void paint(int d){
	int times = 0;
	while (d % 4 == 0) d /= 4, times++;
	
	if (d % 2 == 1){
		for (int i = 0; i < 2 * n; i++)
			for (int j = 0; j < 2 * n; j++)
				if (((i >> times) + (j >> times)) % 2 == 1) a[i][j] = 1;
	}
	if (d % 2 == 0){
		for (int i = 0; i < 2 * n; i++){
			if ((i >> times) % 2 == 1)
				for (int j = 0; j < 2 * n; j++) a[i][j] = 1;
		}
	}
}

int main(){
	scanf("%d%d%d", &n, &d1, &d2);
	int cnt = 0;
	paint(d1); paint(d2);
	
	for (int i = 0; i < 2 * n; i++){
		for (int j = 0; j < 2 * n; j++){
			if (cnt < n * n && !a[i][j]){
				printf("%d %d\n", i, j);
				cnt++;
			}
		}
	}
	return 0;
}