【BZOJ1228】【SDOI2009】E&D

【题目链接】

• 点击打开链接
  

【思路要点】

• 补档博客，无题解。

【代码】

#include<bits/stdc++.h>
using namespace std;
#define MAXN	1005
int get(int x, int y) {
	if (x % 2 == 1 && y % 2 == 1) return 0;
	else return get((x + 1) / 2, (y + 1) / 2) + 1;
}
int main() {
	int T;
	scanf("%d", &T);
	while (T--) {
		int n, ans = 0;
		scanf("%d", &n);
		for (int i = 1; i <= n / 2; i++) {
			int x, y;
			scanf("%d%d", &x, &y);
			ans ^= get(x, y);
		}
		if (ans) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}
/* Use the program below to print a form of the sg function (try to replace the 4 with 1, 2, 8 ...)
#include<bits/stdc++.h>
using namespace std;
#define MAXN	1005
int sg[MAXN][MAXN];
int mark[MAXN * MAXN];
int n, q, num;
int main() {
	freopen("list.txt", "w", stdout);
	cin >> n;
	for (int s = 1; s <= 2 * n - 1; s++)
	for (int i = 1, j = s - 1; i <= n; i++, j--) {
		if (j > n || j <= 0) continue;
		num++;
		if (i != 1) {
			for (int k = 1; k < i; k++)
				mark[sg[k][i - k]] = num;
		}
		if (j != 1) {
			for (int k = 1; k < j; k++)
				mark[sg[k][j - k]] = num;
		}
		int ans = 0;
		while (mark[ans] == num) ans++;
		sg[i][j] = ans;
	}
	for (int i = 4; i <= n; i += 4) {
		for (int j = 4; j <= n; j += 4)
			printf("%4d", sg[i][j]);
		printf("\n");
	}
	return 0;
}
*/