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December 28, 2016 20:55
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Hackerrank - week code 27 - Hackonacci matrix rotation - study C++ code - Google employee's code
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#include <iostream> | |
#include <cassert> | |
#include <cstdio> | |
#include <math.h> | |
#include <iostream> | |
#include <algorithm> | |
#include <vector> | |
#include <time.h> | |
#include <set> | |
#define ll long long | |
using namespace std; | |
bool matpow(int numdims, ll N) | |
{ | |
if (N%7==4) { | |
return 1; | |
} | |
if (N%7==5) { | |
return 1; | |
} | |
if (N%7==6) { | |
return 0; | |
} | |
if (N%7==0) { | |
return 1; | |
} | |
if (N%7==1) { | |
return 0; | |
} | |
if (N%7==2) { | |
return 0; | |
} | |
if (N%7==3) { | |
return 1; | |
} | |
return false; | |
} | |
bool GetNumber(ll x) { | |
if (x==1) | |
return 1; | |
if (x==2) | |
return 0; | |
if (x==3) | |
return 1; | |
return matpow(3, x-3); | |
} | |
bool A[2000][2000]; | |
bool B[2000][2000]; | |
bool tmp2[2000][2000]; | |
int ans[4]; | |
void Count(int n) { | |
for(int i=0;i<n;i++) | |
for(int j=0;j<n;j++) | |
B[i][j]=A[i][j]; | |
ans[0]=0; | |
for(int rot=1;rot<=3;rot++) { | |
for(int i=0;i<n;i++) | |
for(int j=0;j<n;j++) | |
tmp2[j][n-1-i]=B[i][j]; | |
for(int i=0;i<n;i++) | |
for(int j=0;j<n;j++) { | |
B[i][j]=tmp2[i][j]; | |
} | |
ans[rot]=0; | |
for(int i=0;i<n;i++) | |
for(int j=0;j<n;j++) | |
ans[rot]+=(B[i][j]!=A[i][j]); | |
} | |
} | |
int main() { | |
int n,Q; | |
cin>>n>>Q; | |
for(ll i=0;i<n;i++) | |
for(ll j=0;j<n;j++) { | |
if (GetNumber((i+1)*(i+1)*(j+1)*(j+1))) | |
A[i][j]=1; | |
else | |
A[i][j]=0; | |
} | |
Count(n); | |
for(int q=1;q<=Q;q++) { | |
int ang; | |
cin>>ang; | |
cout << ans[(ang%360)/90] << endl; | |
} | |
} |
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