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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#333844 | #4325. Kraljice | ivanj | 0 | 0ms | 0kb | C++14 | 2.1kb | 2024-02-20 17:52:45 | 2024-02-20 17:52:48 |
answer
#include<bits/stdc++.h>
#define pb push_back
#define x first
#define y second
#define all(a) (a).begin(), (a).end()
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
const int maxn = 1025;
int n;
int r[maxn];
int c[maxn];
int d1[maxn];
int d2[maxn];
int mat[maxn][maxn];
vector<ii> ans;
int check(int x, int y) {
return (r[x] + c[y] + d1[x + y] + d2[x - y + n] + 1) % 2;
}
int play(int x, int y) {
if(!check(x, y)) {
cout << "err\n";
exit(0);
}
ans.pb({x, y});
mat[x][y] = 1;
r[x]++, c[y]++, d1[x + y]++, d2[x - y + n]++;
}
void print() {
for(int i = 0;i < n;i++, cout << "\n")
for(int j = 0;j < n;j++) {
if(!mat[i][j]) cout << (r[i] + c[j] + d1[i + j] + d2[i - j + n]) % 2 << " ";
else cout << "X ";
}
cout << "..................................\n";
}
void resi4() {
play(0, 1);
play(1, 3);
play(0, 2);
play(1, 0);
play(1, 1);
play(2, 3);
play(0, 3);
play(2, 0);
play(2, 1);
play(2, 2);
play(3, 0);
play(3, 2);
play(3, 1);
play(3, 3);
}
void resi3() {
play(1, 2);
play(2, 0);
play(1, 1);
play(0, 0);
play(2, 2);
play(2, 1);
play(0, 1);
play(0, 2);
play(1, 0);
}
void rek(int n) {
if(n <= 2) {play(0, 0);return;}
if(n == 3) {resi3();return;}
if(n == 4) {resi4();return;}
rek(n - 2);
for(int i = 0;i < n - 3;i++) {
if(check(i, n - 1))
play(i, n - 1), play(i, n - 2);
else play(i, n - 2), play(i, n - 1);
if(check(n - 1, i))
play(n - 1, i), play(n - 2, i);
else play(n - 2, i), play(n - 1, i);
}
if(n % 2 == 1) {
play(n - 2, n - 1);
play(n - 3, n - 2);
play(n - 1, n - 2);
play(n - 2, n - 2);
play(n - 3, n - 1);
play(n - 1, n - 3);
play(n - 2, n - 3);
play(n - 1, n - 1);
} else {
play(n - 3, n - 1);
play(n - 3, n - 2);
play(n - 2, n - 1);
play(n - 2, n - 3);
play(n - 1, n - 3);
play(n - 2, n - 2);
play(n - 1, n - 2);
play(n - 1, n - 1);
}
}
int main() {
scanf("%d", &n);
rek(n);
printf("%d\n", (int)ans.size());
for(ii p : ans)
printf("%d %d\n", p.x + 1, p.y + 1);
return 0;
}
詳細信息
Subtask #1:
score: 0
Runtime Error
Test #1:
score: 0
Runtime Error
input:
1
output:
result:
Subtask #2:
score: 0
Skipped
Dependency #1:
0%
Subtask #3:
score: 0
Skipped
Dependency #1:
0%
Subtask #4:
score: 0
Skipped
Dependency #1:
0%