Mine Sweeper

Mine Sweeper

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Special Judge

Problem Description

A mine-sweeper map can be expressed as an $r\times c$ grid, and each cell of the grid is either mine cell or non-mine cell. A mine cell has no number on it, while a non-mine cell has a number, the number of mine cells that share at least one common point with the cell, on it. Following is a $16\times 30$ mine-sweeper map, where flag cells denotes mine cells while blank cells denotes non-mine cells with number $0$ on them.

Given an integer $S$, construct a mine-sweeper map of $r,c$ both not exceeding $25$, whose sum of numbers on non-mine cells exactly equals $S$. If multiple solutions exist, print any one of them. If no solution, print -1 in one line.

Input

The first line contains one positive integer $T(1\le T\le 1001)$, denoting the number of test cases. For each test case:

Input one line containing one integer $S(0\le S\le 1000)$.

Output

For each test case:

If no solution, print -1 in one line.

If any solution exists, print two integers $r,c(1\le r,c\le 25)$ in the first line, denoting the size of the mine-sweeper map. Following r lines each contains a string only containing . or X of length c where ., X denote non-mine cells and mine cells respectively, denoting each row of the mine-sweeper map you construct.

Please notice that you needn’t print the numbers on non-mine cells since these numbers can be determined by the output mine-sweeper map.

Sample Input

2
7
128

Sample Output

2 4
X..X
X...
5 19
.XXXX..XXXXX..XXXX.
XX.......X....X..XX
X........X....XXXX.
XX.....X.X....X..XX
.XXXX...XX....XXXX.

Hint

For the first test case, the map with numbers is as follows:

X21X
X211

The sum of these numbers equals $2+1+2+1+1=7$.

思路

1. 如果 $S\le 24$，我们可以构造这样的地图："$.X.X.X\cdots$"，可知当长度为 $l$ 的时候，数字和就等于$l-1$。
2. 如果 $S>24$，我们可以把 $S$ 写成 $S=8a+3b$ 的形式，其中 $a,b>0;b<8$，$(gcd(a,b)=1那么不能组成最大的数为a\ast b-a-b)$那么我们可以构造类
   似如下的地图：

. . . . . . . . .
. X . X . X . X .
. . . . . . . . .
. X . X . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . X X

其中包含恰好 $a$ 个孤立的 X 和 $b$ 个右下角那种连续的 X，可知一个孤立的X 可以对数字和 带来 $8$ 的贡献，而右下角那种连续的X一个可以带来 $3$ 的贡献。

/***  Amber  ***/
#pragma GCC optimize(3,"Ofast","inline")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define ls (rt<<1)
#define rs (rt<<1|1)
typedef long long ll;
template <typename T>
inline void read(T &x) {
    x = 0;
    static int p;
    p = 1;
    static char c;
    c = getchar();
    while (!isdigit(c)) {
        if (c == '-')p = -1;
        c = getchar();
    }
    while (isdigit(c)) {
        x = (x << 1) + (x << 3) + (c - 48);
        c = getchar();
    }
    x *= p;
}
template <typename T>
inline void print(T x) {
    if (x<0) {
        x = -x;
        putchar('-');
    }
    static int cnt;
    static int a[50];
    cnt = 0;
    do {
        a[++cnt] = x % 10;
        x /= 10;
    } while (x);
    for (int i = cnt; i >= 1; i--)putchar(a[i] + '0');
    puts("");
}
const double pi=acos(-1);
const double eps=1e-6;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
const int maxn = 10010;
int S;
inline void work() {
    read(S);
    if (S < 24) {
        printf("1 %d\n", S + 1);
        if (S & 1) {
            for (int i = 1; i <= S / 2 + 1; i++) {
                printf(".X");
            }
        } else {
            for (int i = 1; i <= S / 2; i++) {
                printf(".X");
            }
            putchar('.');
        }
        puts("");
        return;
    }
    int a = S / 8, b = 0;
    while (1) {
        if ((S - a * 8) % 3 == 0) {
            b = (S - a * 8) / 3;
            break;
        }
        a--;
    }
    printf("25 25\n");
    for (int i = 1; i < 25; i++) {
        if (i & 1) {
            for (int j = 1; j <= 25; j++) {
                putchar('.');
            }
        } else {
            for (int j = 1; j <= 25; j++) {
                if (j % 2 == 0 && a > 0) {
                    putchar('X');
                    a--;
                } else putchar('.');
            }
        }
        puts("");
    }
    for (int i = 1; i <= 25; i++) {
        if (i > 25 - b) putchar('X'); else putchar('.');
    }
    puts("");
}
int main() {
    //freopen("1.txt","r",stdin);
    int T = 1;
    read(T);
    while (T--) {
        work();
    }
    return 0;
}