# Fairy Chess

### Problem Statement :

```Let's play Fairy Chess!

You have an n*n chessboard. An s-leaper is a chess piece which can move from some square (x0,y0) to some square (x1,y1) if abs(x0-x1) + abs(y0-y1) <= s; however, its movements are restricted to up, down, left, and right within the confines of the chessboard, meaning that diagonal moves are not allowed. In addition, the leaper cannot leap to any square that is occupied by a pawn.

Given the layout of the chessboard, can you determine the number of ways a leaper can move m times within the chessboard?

Note: abs(x) refers to the absolute value of some integer, x.

Input Format

The first line contains an integer, q, denoting the number of queries. Each query is described as follows:

1.The first line contains three space-separated integers denoting n, m, and s, respectively.
2.Each line i of the n subsequent lines contains n characters. The jth character in the ith line describes the contents of square (i,j) according to the following key:
. indicates the location is empty.
P indicates the location is occupied by a pawn.
L indicates the location of the leaper.
Constraints

1 <= q <= 10
1 <= m <= 200
There will be exactly one L character on the chessboard.
The s-leaper can move up (), down (), left (), and right () within the confines of the chessboard. It cannot move diagonally.```

### Solution :

```                            ```Solution in C :

In C++ :

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cassert>
using namespace std;

const int P = 1000000007;
const int maxN = 200;
const int maxS = 200;

int TN, TC;

int N, M, S, K;

char tmp[maxN + 1];
char board[maxN][maxN];

int sx, sy;

int num[maxN][maxN];
int sum[2][maxN][maxN];

#define PLUS(x, v) \
{ \
x += v; \
if (x >= P) \
x -= P; \
}

#define MINUS(x, v) \
{ \
x -= v; \
if (x < 0) \
x += P; \
}

void find_leaper (int &px, int &py)
{
for (int x = 0; x < N; ++x)
for (int y = 0; y < N; ++y)
if (board[x][y] == 'L')
{
px = x;
py = y;
return;
}
}

void coor_trans (int x, int y, int &tx, int &ty, int &s)
{
if (!((x + y) & 1))
{
s = 0;
tx = (x + y) >> 1;
ty = ((y - x) >> 1) + ((N - 1) >> 1);
}
else
{
s = 1;
tx = (x + y) >> 1;
ty = ((y - x + 1) >> 1) + ((N - 2) >> 1);
}
}

int square_sum (int x, int y, int h)
{
int tx, ty, s;
coor_trans(x, y, tx, ty, s);
int tx2 = tx - h, ty2 = ty - h;
if (tx >= N)
tx = N - 1;
if (ty >= N)
ty = N - 1;
if (tx2 >= N)
tx2 = N - 1;
if (ty2 >= N)
ty2 = N - 1;
int r = sum[s][tx][ty];
if (tx2 >= 0)
MINUS(r, sum[s][tx2][ty]);
if (ty2 >= 0)
MINUS(r, sum[s][tx][ty2]);
if (tx2 >= 0 && ty2 >= 0)
PLUS(r, sum[s][tx2][ty2]);
return r;
}

int solve ()
{
for (int x = 0; x < N; ++x)
for (int y = 0; y < N; ++y)
if (board[x][y] != 'P')
num[x][y] = 1;
else
num[x][y] = 0;

for (int u = 0; u < M; ++u)
{
memset(sum, 0, sizeof(sum));
for (int x = 0; x < N; ++x)
for (int y = 0; y < N; ++y)
if (board[x][y] != 'P')
{
int tx, ty, s;
coor_trans(x, y, tx, ty, s);
sum[s][tx][ty] = num[x][y];
}

for (int s = 0; s < 2; ++s)
{
for (int y = 1; y < N; ++y)
PLUS(sum[s][0][y], sum[s][0][y - 1]);
for (int x = 1; x < N; ++x)
{
PLUS(sum[s][x][0], sum[s][x - 1][0]);
for (int y = 1; y < N; ++y)
{
int &r = sum[s][x][y];
PLUS(r, sum[s][x - 1][y]);
PLUS(r, sum[s][x][y - 1]);
MINUS(r, sum[s][x - 1][y - 1]);
}
}
}

for (int x = 0; x < N; ++x)
for (int y = 0; y < N; ++y)
if (board[x][y] != 'P')
{
num[x][y] = square_sum(x, y + S, S + 1) + square_sum(x, y + S - 1, S);
if (num[x][y] >= P)
num[x][y] -= P;
}
}

return num[sx][sy];
}

int main ()
{
scanf("%d", &TN);
for (TC = 1; TC <= TN; ++TC)
{
scanf("%d%d%d ", &N, &M, &S);
K = N + (S + 1) / 2;
for (int x = 0; x < N; ++x)
{
gets(tmp);
memcpy(board[x], tmp, N);
}
find_leaper(sx, sy);
int ans = solve();
printf("%d\n", ans);
}
}

In Java :

import java.util.Scanner;

public class Solution {

private static final int MOD_PRIME = 1000000007;

public static int addModulo(int x, int y) {
int s = x + y;
if (s >= MOD_PRIME)
s -= MOD_PRIME;
return s;
}

public static int subtractModulo(int x, int y) {
int d = x - y;
if (d < 0)
d += MOD_PRIME;
return d;
}

public static int query(int[][] sum, int x, int y, int s) {
int n = sum.length;
int rx = n / 2 + x - y;
int ry = x + y + 1;
int high_x = Math.min(rx + s, n - 1);
int low_x = Math.max(rx - s, 1);
int high_y = Math.min(ry + s, n - 1);
int low_y = Math.max(ry - s, 1);
int pos = addModulo(sum[high_x][high_y], sum[low_x - 1][low_y - 1]);
int neg = addModulo(sum[high_x][low_y - 1], sum[low_x - 1][high_y]);
return subtractModulo(pos, neg);
}

public static void createSumTable(int[][] sum, int[][] matrix) {
for (int i = 0; i < matrix.length; i++) {
int partialSum = 0;
for (int j = 0; j < matrix[i].length; j++) {
sum[i][j] = partialSum;
if (i > 0)
sum[i][j] = addModulo(sum[i][j], sum[i - 1][j]);
}
}
}

public static final void solveProblem(Scanner in) {
String line = in.nextLine();
String tokens[] = line.split(" ");
int n = Integer.parseInt(tokens[0]);
int m = Integer.parseInt(tokens[1]);
int s = Integer.parseInt(tokens[2]);

int[][] rotated = new int[2 * n][2 * n];
int[][] sum = new int[2 * n][2 * n];
int[][] next = new int[2 * n][2 * n];

int lx = -1, ly = -1;
for (int i = 0; i < n; i++) {
line = in.nextLine();
for (int j = 0; j < line.length(); j++) {
char ch = line.charAt(j);
if (ch != 'P') {
rotated[n + i - j][i + j + 1] = 1;
if (ch == 'L') {
lx = i;
ly = j;
}
}
}
}
createSumTable(sum, rotated);

for (int k = 2; k <= m; k++) {
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) {
int x = n + i - j;
int y = i + j + 1;
if (rotated[x][y] != 0)
next[x][y] = query(sum, i, j, s);
}
int[][] tmp = next;
next = rotated;
rotated = tmp;
createSumTable(sum, rotated);
}

System.out.println(query(sum, lx, ly, s));
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);

int numTests = Integer.parseInt(in.nextLine());
for (int i = 0; i < numTests; i++)
solveProblem(in);

in.close();
}

}

In C :

#include <stdio.h>

#define P 1000000007

long long ll,t,a[2][1000][1000],mm,mmm,p[1010][1010];
long long r[1010][1010],i,j,k,l,m,n;
char s[210][210];

/*
long long bb(long long ll, long long ii, long long jj)
{

if(ii<0 || ii>=n || jj<0 || jj>=n) return 0;

return a[ll][ii][jj];
}
*/

int main()
{

scanf("%lld",&t);
while(t--)
{
scanf("%lld %lld %lld\n",&n,&m,&k);
for(i=0;i<n;i++) scanf("%s\n",s[i]);

/*
for(i=0;i<n;i++)
for(j=0;j<n;j++)
for(l=0;l<0;l++) a[l][i][j]=0;
*/

for(i=0;i<1000;i++)
for(j=0;j<1000;j++)
a[0][i][j] = a[1][i][j] = 0;

for(i=0;i<n;i++)
for(j=0;j<n;j++)
if(s[i][j]=='L') a[0][i+500][j+500] = 1;

for(i=0;i<1000;i++)
for(j=0;j<1000;j++) p[i][j]= r[i][j] =0;

//for(l=0;l<m;l++)
l=0;
ll=1;

while(m--)
{

for(i=500;i<n+500;i++)
{
for(j=-k+500;j<n+500;j++)
{
p[i][j] = (p[i-1][j+1] + a[l][i][j] - a[l][i-1-k][j+1+k])%P;
}
}

for(i=n+500;i<n+k+500;i++)
{
for(j=500;j<n+500;j++)
{
p[i][j] = (p[i-1][j+1] + a[l][i][j] - a[l][i-1-k][j+1+k])%P;
}
}

for(i=n-1+500;i>=500-1;i--)
{
for(j=-k+500;j<=n+500;j++)
{
r[i][j] = (r[i+1][j+1] + a[l][i][j] - a[l][i+1+k][j+1+k])%P;
}
}

for(i=-2+500;i>=-k+500;i--)
{
for(j=500;j<=n+500;j++)
{
r[i][j] = (r[i+1][j+1] + a[l][i][j] - a[l][i+1+k][j+1+k])%P;
}
}

j = 500-1;
//jj = 500+j;
mmm=0;
for(i=-k+500;i<500;i++)
{
//   ii = i+500;
mmm = (mmm + p[i+k][j])%P;
//+ r[i][j-k]
//- a[l][i+k][j]
//- p[i-1][j-k] - r[i-1-k][j]
//+ a[l][i-1-k][j])%P;

}

for(j=500;j<n+500;j++)
{
//     jj = j+500;
i = 500-1;
//     ii= 500+i;

mmm = (mmm + p[i+k][j] + r[i-k][j] - a[l][i][j+k]
- p[i][j-1-k] - r[i][j-1-k] + a[l][i][j-k-1])%P;

mm = mmm;

for(i=500;i<n+500;i++)
{
//       ii = i+500;
mm = (mm + p[i+k][j] + r[i][j-k] - a[l][i+k][j]
- p[i-1][j-k] - r[i-1-k][j] + a[l][i-1-k][j])%P;

//  printf("mm %lld %lld %lld\n",i,j,mm);

if(s[i-500][j-500]!='P') a[ll][i][j] = mm;
else a[ll][i][j] = 0;

}
}

l = (l+1)%2;
ll = (ll+1)%2;

/*
for(i=500;i<n+500;i++)
{
for(j=500;j<n+500;j++) printf("%lld ",a[l][i][j]);
printf("\n");
}

printf("--------------\n");
*/
}

mm = 0;
for(i=500;i<n+500;i++)
for(j=500;j<n+500;j++) mm = (mm + a[l][i][j])%P;

printf("%lld\n",(mm+P)%P);

//return 0;

}

return 0;
}```
```

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