이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MAXN = 70;
int K, N;
char ans[MAXN][MAXN];
int solve(char a[MAXN][MAXN], int N) {
int dp[MAXN][MAXN][2];
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
dp[i][j][0] = dp[i][j][1] = 0;
dp[0][0][0] = 1;
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (i == N-1 && j == N-1) break;
if (a[i][j] == 'X') {
dp[i+1][j][0] += dp[i][j][0] + dp[i][j][1];
dp[i][j+1][1] += dp[i][j][0] + dp[i][j][1];
} else if (a[i][j] == '.') {
dp[i+1][j][0] += dp[i][j][0];
dp[i][j+1][1] += dp[i][j][1];
} else if (a[i][j] == 'r') {
dp[i][j+1][1] += dp[i][j][0] + dp[i][j][1];
} else if (a[i][j] == 'd') {
dp[i+1][j][0] += dp[i][j][0] + dp[i][j][1];
}
}
}
return dp[N-1][N-1][0] + dp[N-1][N-1][1];
}
void mkarr(int K, int N) {
for (int i = 0; i < N; i++) {
ans[i][N-1] = 'd';
ans[N-1][i] = 'r';
}
ans[N-1][N-1] = '.';
for (int i = 0; i+1 < N; i++) {
for (int j = 0; j+1 < N; j++) {
if (i == j)
ans[i][j] = 'X';
else if (i+1 == j)
ans[i][j] = 'd';
else if (j+1 == i)
ans[i][j] = 'r';
else
ans[i][j] = '.';
}
}
if (K != (1LL<<(N-1))) {
ans[0][1] = ans[1][0] = 'X';
int c = -1, sum = (1LL<<(N-1));
for (int j = 2; j+1 < N; j++) {
if ((1LL<<(N-j)) & K) {
c = j, ans[0][j] = ans[j][0] = 'X';
sum += (1LL<<(N-j));
}
}
sum += 2;
if (sum >= K + 1 && c != -1) ans[0][c] = 'd';
if (sum >= K + 2 && c != -1) ans[c][0] = 'r';
//cout << sum << "\n";
}
/*
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
cout << ans[i][j];
}
cout << "\n";
}
*/
}
int lb(int a) {
return a &- a;
}
signed main() {
cin >> K;
int t = K;
while (t != lb(t)) t -= lb(t);
N = __lg(t*2);
mkarr(K, N);
if (K == 3) {
N = 3;
ans[0][0] = 'X';
ans[1][0] = 'X';
ans[2][0] = ans[0][1] = ans[1][1] = '.';
ans[2][0] = ans[2][1] = 'r';
ans[0][2] = ans[1][2] = 'd';
ans[2][2] = '.';
}
cout << N << " " << N << "\n";
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
cout << ans[i][j];
}
cout << "\n";
}
//cout << solve(ans, N) << "\n";
return 0;
}
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