Here's a cube whose size of its 3 dimensions are all infinite. Meanwhile, there're 6 programs operating this cube:
FILL(X,Y,Z): Fill some part of the cube with different values.
memset(cube, 0, sizeof(cube));
puts("START");
cnt = 0;
for (int i = 0; i < X; i++) {
for (int j = 0; j < Y; j++) {
for (int k = 0; k < Z; k++) {
cube[i][j][k] = ++cnt;
}
}
}
SWAP1(x1,x2): Swap two sub-cube along the first dimensions.
for (int j = 0; j < Y; j++) {
for (int k = 0; k < Z; k++) {
exchange(cube[x1][j][k], cube[x2][j][k]);
}
}
SWAP2(y1,y2): Swap two sub-cube along the second dimensions.
for (int i = 0; i < X; i++) {
for (int k = 0; k < Z; k++) {
exchange(cube[i][y1][k], cube[i][y2][k]);
}
}
SWAP3(z1,z2): Swap two sub-cube along the third dimensions.
for (int i = 0; i < X; i++) {
for (int j = 0; j < Y; j++) {
exchange(cube[i][j][z1], cube[i][j][z2]);
}
}
FIND(value): Output the value's location, if it exist.
for (int i = 0; i < X; i++) {
for (int j = 0; j < Y; j++) {
for (int k = 0; k < Z; k++) {
if (cube[i][j][k] == value) {
printf("%d %d %d\n", i, j, k);
}
}
}
}
QUERY(x,y,z): Output the value at (x,y,z).
printf("%d\n", cube[x][y][z]);
We'll give a list of operations mentioned above. Your job is to simulate the program and tell us what does the machine output in progress.
Input
There'll be 6 kind of operations in the input.
FILL X Y Z (1 <= X, Y, Z <= 1000) for FILL(X,Y,Z)
SWAP1 X1 X2 (0 <= X1, X2 < X) for SWAP1(X1,X2)
SWAP2 Y1 Y2 (0 <= Y1, Y2 < Y) for SWAP2(Y1,Y2)
SWAP3 Z1 Z2 (0 <= Z1, Z2 < Z) for SWAP3(Z1,Z2)
FIND value (value > 0) for FIND(value)
QUERY x y z (0 <= x < X, 0 <= y < Y, 0 <= z < Z) for QUERY(x,y,z)
The input will always start with FILL operation and terminate by EOF.
The number of the operations will less than 200,000, while the FILL operation will less than 100.
Output
Simulate all of the operations in order, and print the output of the programs.
Sample Input
FILL 2 3 1
SWAP1 0 1
SWAP2 0 2
SWAP3 0 0
FIND 1
FIND 2
FIND 3
FIND 4
FIND 5
FIND 6
FIND 7
QUERY 0 0 0
QUERY 0 1 0
QUERY 0 2 0
QUERY 1 0 0
QUERY 1 1 0
QUERY 1 2 0
Sample Output
START
1 2 0
1 1 0
1 0 0
0 2 0
0 1 0
0 0 0
6
5
4
3
2
1
