「CF400C」Inna and Huge Candy Matrix
Inna and Dima decided to surprise Sereja. They brought a really huge candy matrix, it’s big even for Sereja! Let’s number the rows of the giant matrix from 1 to n from top to bottom and the columns — from 1 to m, from left to right. We’ll represent the cell on the intersection of the i-th row and j-th column as (i, j). Just as is expected, some cells of the giant candy matrix contain candies. Overall the matrix has p candies: the k-th candy is at cell (xk, yk).
The time moved closer to dinner and Inna was already going to eat p of her favourite sweets from the matrix, when suddenly Sereja (for the reason he didn’t share with anyone) rotated the matrix x times clockwise by 90 degrees. Then he performed the horizontal rotate of the matrix y times. And then he rotated the matrix z times counterclockwise by 90 degrees. The figure below shows how the rotates of the matrix looks like.
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The first line of the input contains fix integers n, m, x, y, z, p (1 ≤ n, m ≤ 109; 0 ≤ x, y, z ≤ 109; 1 ≤ p ≤ 105).
Each of the following p lines contains two integers xk, yk (1 ≤ xk ≤ n; 1 ≤ yk ≤ m) — the initial coordinates of the k-th candy. Two candies can lie on the same cell.
For each of the p candies, print on a single line its space-separated new coordinates.
1 2 3 4 5 6 7 8 9 10 |
3 3 3 1 1 9 1 1 1 2 1 3 2 1 2 2 2 3 3 1 3 2 3 3 |
1 2 3 4 5 6 7 8 9 |
1 3 1 2 1 1 2 3 2 2 2 1 3 3 3 2 3 1 |
Just for clarity. Horizontal rotating is like a mirroring of the matrix. For matrix:
1 2 3 |
QWER REWQ ASDF -> FDSA ZXCV VCXZ |
看了眼数据范围。非常激动地写了个矩阵乘法。其实把xyz取模就。。。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 |
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; int n,m,a1,a2,a3,p; int b[5][5],r[5][5],l[5][5],f[5][5]; void mul(int a[5][5],int b[5][5],int ans[5][5]) { int t[5][5]; for(int i=1;i<5;i++) for(int j=1;j<5;j++) { t[i][j]=0; for(int k=1;k<5;k++) t[i][j]+=a[i][k]*b[k][j]; } for(int i=1;i<5;i++) for(int j=1;j<5;j++) ans[i][j]=t[i][j]; } void mul2(int a[5],int b[5][5],int ans[5]) { int t[5]; for(int j=1;j<5;j++) { t[j]=0; for(int k=1;k<5;k++) t[j]+=a[k]*b[k][j]; } for(int j=1;j<5;j++) ans[j]=t[j]; } int main() { scanf("%d%d%d%d%d%d",&n,&m,&a1,&a2,&a3,&p); b[1][1]=b[2][2]=b[3][3]=b[4][4]=1; r[2][1]=1;r[3][2]=1;r[1][2]=-1;r[4][3]=1;r[3][4]=1; f[1][1]=1;f[4][2]=1;f[2][2]=-1;f[3][3]=1;f[4][4]=1; l[2][1]=-1;l[4][1]=1;l[1][2]=1;l[4][3]=1;l[3][4]=1; while(a1) { if(a1&1)mul(b,r,b); a1>>=1; mul(r,r,r); } while(a2) { if(a2&1)mul(b,f,b); a2>>=1; mul(f,f,f); } while(a3) { if(a3&1)mul(b,l,b); a3>>=1; mul(l,l,l); } int a[5]; for(int i=1;i<=p;i++) { scanf("%d%d",&a[1],&a[2]); a[3]=n+1;a[4]=m+1; mul2(a,b,a); printf("%d %d\n",a[1],a[2]); } return 0; } |