King Richard's Knights
Problem Statement :
King Richard is leading a troop of knights into battle! Being very organized, he labels his knights and arranges them in an square formation, demonstrated below: Before the battle begins, he wants to test how well his knights follow instructions. He issues drill commands, where each command follows the format ai bi di and is executed like so: All knights in the square having the top-left corner at location and the bottom-right corner at location rotate in the clockwise direction. Recall that some location denotes the cell located at the intersection of row and column . For example: You must follow the commands sequentially. The square for each command is completely contained within the square for the previous command. Assume all knights follow the commands perfectly. After performing all drill commands, it's time for battle! King Richard chooses knights for his first wave of attack; however, because the knights were reordered by the drill commands, he's not sure where his chosen knights are! As his second-in-command, you must find the locations of the knights. For each knight , , print the knight's row and column locations as two space-separated values on a new line. Input Format This is broken down into three parts: The first line contains a single integer, . The second line contains a single integer, . Each line of the subsequent lines describes a command in the form of three space-separated integers corresponding to , , and , respectively. The next line contains a single integer, . Each line of the subsequent lines describes a knight the King wants to find in the form of a single integer Output Format Print lines of output, where each line contains two space-separated integers describing the respective row and column values where knight is located.
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <inttypes.h>
#include <string.h>
int main()
{
int64_t N, S, L;
int64_t * command, *initialArea, *searchArea;
scanf ( "%lld", &N );
scanf ( "%lld", &S );
command = ( int64_t * ) malloc ( 3 * S * sizeof ( int64_t ) );
initialArea = ( int64_t * ) malloc ( 3 * S * sizeof ( int64_t ) );
searchArea = ( int64_t * ) malloc ( 4 * S * sizeof ( int64_t ) );
for ( int64_t i = 0; i < S; i++ )
{
scanf ( "%lld%lld%lld", command + i * 3, command + i * 3 + 1, command + i * 3 + 2 );
command[3 * i + 0]--;
command[3 * i + 1]--;
}
initialArea[0] = command[0];
initialArea[1] = command[1];
initialArea[2] = command[2];
searchArea[4 * 0 + 0] = initialArea[3 * 0 + 0];
searchArea[4 * 0 + 1] = initialArea[3 * 0 + 0] + initialArea[3 * 0 + 2];
searchArea[4 * 0 + 2] = initialArea[3 * 0 + 1];
searchArea[4 * 0 + 3] = initialArea[3 * 0 + 1] + initialArea[3 * 0 + 2];
for ( int64_t i = 1; i < S; i++ )
{
int64_t di = command[3 * i + 0] - command[3 * ( i - 1 ) + 0];
int64_t dj = command[3 * i + 1] - command[3 * ( i - 1 ) + 1];
initialArea[3 * i + 2] = command[3 * i + 2];
switch ( i % 4 )
{
case 0:
initialArea[3 * i + 0] = initialArea[3 * ( i - 1 ) + 0] + di;
initialArea[3 * i + 1] = initialArea[3 * ( i - 1 ) + 1] + dj;
break;
case 1:
initialArea[3 * i + 0] = initialArea[3 * ( i - 1 ) + 0] - dj + initialArea[3 * ( i - 1 ) + 2] - initialArea[3 * i + 2];
initialArea[3 * i + 1] = initialArea[3 * ( i - 1 ) + 1] + di;
break;
case 2:
initialArea[3 * i + 0] = initialArea[3 * ( i - 1 ) + 0] - di + initialArea[3 * ( i - 1 ) + 2] - initialArea[3 * i + 2];
initialArea[3 * i + 1] = initialArea[3 * ( i - 1 ) + 1] - dj + initialArea[3 * ( i - 1 ) + 2] - initialArea[3 * i + 2];
break;
case 3:
initialArea[3 * i + 0] = initialArea[3 * ( i - 1 ) + 0] + dj;
initialArea[3 * i + 1] = initialArea[3 * ( i - 1 ) + 1] - di + initialArea[3 * ( i - 1 ) + 2] - initialArea[3 * i + 2];
break;
}
searchArea[4 * i + 0] = initialArea[3 * i + 0];
searchArea[4 * i + 1] = initialArea[3 * i + 0] + initialArea[3 * i + 2];
searchArea[4 * i + 2] = initialArea[3 * i + 1];
searchArea[4 * i + 3] = initialArea[3 * i + 1] + initialArea[3 * i + 2];
}
// for ( int64_t i = 0; i < S; i++ )
// {
// printf ( "(%d, %d, %d) (%d, %d, %d)\n", command[3 * i + 0], command[3 * i + 1], command[3 * i + 2], initialArea[3 * i + 0], initialArea[3 * i + 1], initialArea[3 * i + 2]);
// }
scanf ( "%lld", &L );
while ( L-- > 0 )
{
int64_t w;
scanf ( "%lld", &w );
int64_t i = w / N, j = w % N;
int64_t max = S - 1, min = -1, last = max / 2;
while ( max > min )
{
if ( i >= searchArea[4 * last + 0] &&
i <= searchArea[4 * last + 1] &&
j >= searchArea[4 * last + 2] &&
j <= searchArea[4 * last + 3] )
{
min = last;
int64_t temp = min + max;
if ( temp % 2 == 1 )
{
last = temp / 2 + 1;
}
else
{
last = temp / 2;
}
}
else
{
max = last - 1;
last = ( min + max ) / 2;
}
}
// printf("last = %lld\n", last);
int64_t di = i - initialArea[3 * last + 0];
int64_t dj = j - initialArea[3 * last + 1];
if ( last >= 0 )
{
switch ( ( last + 1 ) % 4 )
{
case 0:
i = di + command[3 * last + 0];
j = dj + command[3 * last + 1];
break;
case 1:
i = dj + command[3 * last + 0];
j = command[3 * last + 1] + command[3 * last + 2] - di;
break;
case 2:
i = command[3 * last + 0] + command[3 * last + 2] - di;
j = command[3 * last + 1] + command[3 * last + 2] - dj;
break;
case 3:
i = command[3 * last + 0] + command[3 * last + 2] - dj;
j = di + command[3 * last + 1];
break;
}
}
printf ( "%lld %lld\n", i + 1, j + 1 );
}
return 0;
}
Solution in C++ :
In C++ :
#include <bits/stdc++.h>
using namespace std;
int S,N,L;
typedef complex<int> pt;
struct state {
pt orig;
pt now;
int d,idx;
pt get_at(int x,int y) {
x-=real(now);
y-=imag(now);
int rots=idx%4;
swap(x,y);
for(;rots--;) {
int tmp=x;
x=y;
y=d-tmp;
}
swap(x,y);
return pt(real(orig)+x,imag(orig)+y);
}
bool has(int x,int y) {
return x>=real(orig) && y>=imag(orig) && x<=real(orig)+d && y<=imag(orig)+d;
}
pt rlook(int x,int y) {
swap(orig,now);
idx = 4-(idx%4);
pt ans=get_at(x,y);
idx=4-(idx%4);
swap(orig,now);
return ans;
}
};
vector<state> vs;
int main() {
cin>>N>>S;
vs.push_back({pt(1,1),pt(1,1),N-1,0});
for(int i=1;i<=S;i++) {
int a,b,d; cin>>a>>b>>d;
pt ul=vs.back().get_at(a,b);
pt dr=vs.back().get_at(a+d,b+d);
pt neworig={min(real(ul),real(dr)), min(imag(ul),imag(dr))};
pt newnow={a,b};
vs.push_back({neworig,newnow,d,i});
}
for(auto &st : vs) {
// cout<<st.orig<<" "<<st.now<<" "<<st.d<<" "<<st.idx<<endl;
}
cin>>L;
for(;L--;) {
long long w; cin>>w;
int x=w/N+1;
int y=w%N+1;
int lb=0,rb=S;
for(;lb<rb;) {
int mb=(lb+rb+1)/2;
if(vs[mb].has(x,y)) lb=mb;
else rb=mb-1;
}
pt p=vs[lb].rlook(x,y);
cout<<real(p)<<" "<<imag(p)<<endl;
}
}
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int s = sc.nextInt();
long[] a = new long[s+1];
long[] b = new long[s+1];
long[] d = new long[s+1];
long[] tln = new long[s+1];
a[0] = 1;
b[0] = 1;
d[0] = n-1;
for (int i = 1; i <= s; i++) {
a[i] = sc.nextInt();
b[i] = sc.nextInt();
d[i] = sc.nextInt();
if (i%4==1) {
tln[i] = tln[i-1]+(a[i]-a[i-1])*n+(b[i]-b[i-1]);
} else if (i%4==2) {
tln[i] = tln[i-1]+((b[i-1]+d[i-1])-(b[i]+d[i]))*n+(a[i]-a[i-1]);
} else if (i%4==3) {
tln[i] = tln[i-1]+((a[i-1]+d[i-1])-(a[i]+d[i]))*n+((b[i-1]+d[i-1])-(b[i]+d[i]));
} else if (i%4==0) {
tln[i] = tln[i-1]+(b[i]-b[i-1])*n+((a[i-1]+d[i-1])-(a[i]+d[i]));
}
}
int l = sc.nextInt();
long[] w = new long[l];
for (int i = 0; i < l; i++) {
w[i] = sc.nextLong();
int low = 0;
int high = s;
while (low != high) {
int mid = (low+high+1)/2;
if (w[i] >= tln[mid] && w[i] < tln[mid]+(d[mid]+1)*n && w[i]%n >= tln[mid]%n && w[i]%n <= (tln[mid]%n)+d[mid])
low = mid;
else
high = mid-1;
}
long off1 = (w[i]-tln[low])/n;
long off2 = (w[i]-tln[low])%n;
if (low%4==0) {
System.out.println((a[low]+off1)+" "+(b[low]+off2));
} else if (low%4==1) {
System.out.println((a[low]+off2)+" "+(b[low]+d[low]-off1));
} else if (low%4==2) {
System.out.println((a[low]+d[low]-off1)+" "+(b[low]+d[low]-off2));
} else if (low%4==3) {
System.out.println((a[low]+d[low]-off2)+" "+(b[low]+off1));
}
}
}
}
Solution in Python :
In Python3 :
def mat_fac(f):
a, b, d = drill[dn]
if f == 1:
return(((a-b, a+b+d), ((0, 1), (-1,0))))
elif f == 2:
return(((2*a+d, 2*b+d), ((-1, 0), (0, -1))))
elif f == 3:
return(((a+b+d, b-a), ((0, -1), (1, 0))))
def mat_mult(fac, pf):
c2, d2 = fac[0]
c1, d1 = pf[0]
k1, k2 = fac[1]
m1, m2 = pf[1]
cnext, dnext = c2+k1[0]*c1+k1[1]*d1, d2+k2[0]*c1+k2[1]*d1
knext1 = (k1[0]*m1[0]+k1[1]*m2[0], k1[0]*m1[1]+k1[1]*m2[1])
knext2 = (k2[0]*m1[0]+k2[1]*m2[0], k2[0]*m1[1]+k2[1]*m2[1])
return(((cnext, dnext), (knext1, knext2)))
def revnew(drn, row, col):
c, d = drill[drn][3][0]
k1, k2 = drill[drn][3][1]
nrow = c + k1[0]*row + k1[1]*col
ncol = d + k2[0]*row + k2[1]*col
return((nrow, ncol))
def contained(x, row, col):
a, b, d, f = drill[x]
if row < a or row > a+d: return(0)
elif col < b or col > b+d: return(0)
else: return(1)
def bin_s(row, col):
lower, upper = 0, len(drill)-1
if not contained(lower, row, col):
return((row, col))
nrow, ncol = revnew(upper, row, col)
if contained(upper, nrow, ncol):
return((nrow, ncol))
while lower < upper:
x = lower+(upper-lower)//2
nrow, ncol = revnew(x, row, col)
if contained(x, nrow, ncol):
if not contained(x+1, nrow, ncol):
return((nrow, ncol))
else:
lower = x + 1; continue
else:
upper = x
nrow, ncol = revnew(lower, row, col)
return((nrow, ncol))
### Read Data
N = int(input())
S = int(input())
drill = [[int(t) for t in input().split()] for i in range(S)]
L = int(input())
trusty_k = [int(input()) for i in range(L)]
# Process drill matrix to get factors
dn = 0
while dn < len(drill):
if drill[dn][2] == 0:
del(drill[dn])
continue
elif dn+3 < len(drill) and drill[dn] == drill[dn+3]:
del(drill[dn:dn+4])
continue
elif dn+2 < len(drill) and drill[dn] == drill[dn+2]:
factor = mat_fac(3)
del(drill[dn:dn+2])
elif dn+1 < len(drill) and drill[dn] == drill[dn+1]:
factor = mat_fac(2)
del(drill[dn])
else:
factor = mat_fac(1)
if dn == 0:
drill[dn] += [factor]
else:
drill[dn] += [mat_mult(factor, drill[dn-1][3])]
dn += 1
# Locate trusty knights
for sn in range(L):
kn = trusty_k[sn]
row, col = kn//N+1, kn%N+1
newpos = bin_s(row, col)
print(newpos[0], newpos[1])
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