Problem Statement

We have a grid with H rows and W columns. The square at the i-th row and the j-th column will be called Square (i,j).

The integers from 1 through H×W are written throughout the grid, and the integer written in Square (i,j) is Ai,j.

You, a magical girl, can teleport a piece placed on Square (i,j) to Square (x,y) by consuming |xi|+|yj| magic points.

You now have to take Q practical tests of your ability as a magical girl.

The i-th test will be conducted as follows:

  • Initially, a piece is placed on the square where the integer Li is written.

  • Let x be the integer written in the square occupied by the piece. Repeatedly move the piece to the square where the integer x+D is written, as long as x is not Ri. The test ends when x=Ri.

  • Here, it is guaranteed that RiLi is a multiple of D.

For each test, find the sum of magic points consumed during that test.

Constraints

  • 1≤H,W≤300
  • 1≤DH×W
  • 1≤Ai,jH×W
  • Ai,jAx,y((i,j)≠(x,y))
  • 1≤Q≤105
  • 1≤LiRiH×W
  • (RiLi) is a multiple of D.

Input

Input is given from Standard Input in the following format:

H W D
A1,1 A1,2 A1,W
:
AH,1 AH,2 AH,W
Q
L1 R1
:
LQ RQ

Output

For each test, print the sum of magic points consumed during that test.

Output should be in the order the tests are conducted.

Sample Input 1

3 3 2
1 4 3
2 5 7
8 9 6
1
4 8

Sample Output 1

5
  • 4 is written in Square (1,2).

  • 6 is written in Square (3,3).

  • 8 is written in Square (3,1).

Thus, the sum of magic points consumed during the first test is (|3−1|+|3−2|)+(|3−3|+|1−3|)=5.

Sample Input 2

4 2 3
3 7
1 4
5 2
6 8
2
2 2
2 2

Sample Output 2

0
0

Note that there may be a test where the piece is not moved at all, and there may be multiple identical tests.

Sample Input 3

5 5 4
13 25 7 15 17
16 22 20 2 9
14 11 12 1 19
10 6 23 8 18
3 21 5 24 4
3
13 13
2 10
13 13

Sample Output 3

0
5
0

Q最高100000,如果每次都要计算,会超时,先记录前缀和,直接用前缀和相减就好了。
代码:
#include <bits/stdc++.h>///int * int = int
using namespace std;
int h,w,d,s[][],q,a,b;
int x[],y[];
int dp[];
int main()
{
scanf("%d%d%d",&h,&w,&d);
for(int i = ;i < h;i ++)
{
for(int j = ;j < w;j ++)
{
scanf("%d",&s[i][j]);
x[s[i][j]] = i;
y[s[i][j]] = j;
}
}
for(int i = + d;i <= h * w;i ++)
{
dp[i] = dp[i - d] + abs(x[i] - x[i - d]) + abs(y[i] - y[i - d]);
}
scanf("%d",&q);
for(int i = ;i < q;i ++)
{
scanf("%d%d",&a,&b);
printf("%lld\n",abs(dp[a] - dp[b]));
}
}

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