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Angled Flip CodeChef Solution
Problem -Angled Flip CodeChef Solution
This website is dedicated for CodeChef solution where we will publish right solution of all your favourite CodeChef problems along with detailed explanatory of different competitive programming concepts and languages.
Angled Flip CodeChef Solution in C++17
#include <bits/stdc++.h>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")//Comment optimisations for interactive problems (use endl)
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
using namespace std;
struct PairHash {inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }};
// speed
#define Code ios_base::sync_with_stdio(false);
#define By ios::sync_with_stdio(0);
#define Sumfi cout.tie(NULL);
// alias
using ll = long long;
using ld = long double;
using ull = unsigned long long;
// constants
const ld PI = 3.14159265358979323846; /* pi */
const ll INF = 1e18;
const ld EPS = 1e-9;
const ll MAX_N = 1010101;
const ll mod = 998244353;
// typedef
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef array<ll,3> all3;
typedef array<ll,5> all5;
typedef vector<all3> vall3;
typedef vector<all5> vall5;
typedef vector<ld> vld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef deque<ll> dqll;
typedef deque<pll> dqpll;
typedef pair<string, string> pss;
typedef vector<pss> vpss;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef unordered_set<ll> usll;
typedef unordered_set<pll, PairHash> uspll;
typedef unordered_map<ll, ll> umll;
typedef unordered_map<pll, ll, PairHash> umpll;
// macros
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,m,n) for(ll i=n;i>=m;i--)
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define INF(a) memset(a,0x3f3f3f3f3f3f3f3fLL,sizeof(a))
#define ASCEND(a) iota(all(a),0)
#define sz(x) ll((x).size())
#define BIT(a,i) (a & (1ll<<i))
#define BITSHIFT(a,i,n) (((a<<i) & ((1ll<<n) - 1)) | (a>>(n-i)))
#define pyes cout<<"YES\n";
#define pno cout<<"NO\n";
#define endl "\n"
#define pneg1 cout<<"-1\n";
#define ppossible cout<<"Possible\n";
#define pimpossible cout<<"Impossible\n";
#define TC(x) cout<<"Case #"<<x<<": ";
#define X first
#define Y second
// utility functions
template <typename T>
void print(T &&t) { cout << t << "\n"; }
template<typename T>
void printv(vector<T>v){ll n=v.size();rep(i,0,n){cout<<v[i];if(i+1!=n)cout<<' ';}cout<<endl;}
template<typename T>
void printvln(vector<T>v){ll n=v.size();rep(i,0,n)cout<<v[i]<<endl;}
void fileIO(string in = "input.txt", string out = "output.txt") {freopen(in.c_str(),"r",stdin); freopen(out.c_str(),"w",stdout);}
void readf() {freopen("", "rt", stdin);}
template<typename T>
void readv(vector<T>& v){rep(i,0,sz(v)) cin>>v[i];}
template<typename T, typename U>
void readp(pair<T,U>& A) {cin>>A.first>>A.second;}
template<typename T, typename U>
void readvp(vector<pair<T,U>>& A) {rep(i,0,sz(A)) readp(A[i]); }
void readvall3(vall3& A) {rep(i,0,sz(A)) cin>>A[i][0]>>A[i][1]>>A[i][2];}
void readvall5(vall5& A) {rep(i,0,sz(A)) cin>>A[i][0]>>A[i][1]>>A[i][2]>>A[i][3]>>A[i][4];}
void readvvll(vvll& A) {rep(i,0,sz(A)) readv(A[i]);}
struct Combination {
vll fac, inv;
ll n, MOD;
ll modpow(ll n, ll x, ll MOD = mod) { if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }
Combination(ll _n, ll MOD = mod): n(_n + 1), MOD(MOD) {
inv = fac = vll(n,1);
rep(i,1,n) fac[i] = fac[i-1] * i % MOD;
inv[n - 1] = modpow(fac[n - 1], MOD - 2, MOD);
rrep(i,1,n - 2) inv[i] = inv[i + 1] * (i + 1) % MOD;
}
ll fact(ll n) {return fac[n];}
ll nCr(ll n, ll r) {
if(n < r or n < 0 or r < 0) return 0;
return fac[n] * inv[r] % MOD * inv[n-r] % MOD;
}
};
struct Matrix {
ll r,c;
vvll matrix;
Matrix(ll r, ll c, ll v = 0): r(r), c(c), matrix(vvll(r,vll(c,v))) {}
Matrix operator*(const Matrix& B) const {
Matrix res(r, B.c);
rep(i,0,r) rep(j,0,B.c) rep(k,0,B.r) {
res.matrix[i][j] = (res.matrix[i][j] + matrix[i][k] * B.matrix[k][j] % mod) % mod;
}
return res;
}
Matrix copy() {
Matrix copy(r,c);
copy.matrix = matrix;
return copy;
}
Matrix pow(ll n) {
assert(r == c);
Matrix res(r,r);
Matrix now = copy();
rep(i,0,r) res.matrix[i][i] = 1;
while(n) {
if(n & 1) res = res * now;
now = now * now;
n /= 2;
}
return res;
}
};
// geometry data structures
template <typename T>
struct Point {
T y,x;
Point(T y, T x) : y(y), x(x) {}
Point(pair<T,T> p) : y(p.first), x(p.second) {}
Point() {}
void input() {cin>>y>>x;}
friend ostream& operator<<(ostream& os, const Point<T>& p) { os<<p.y<<' '<<p.x<<'\n'; return os;}
Point<T> operator+(Point<T>& p) {return Point<T>(y + p.y, x + p.x);}
Point<T> operator-(Point<T>& p) {return Point<T>(y - p.y, x - p.x);}
Point<T> operator*(ll n) {return Point<T>(y*n,x*n); }
Point<T> operator/(ll n) {return Point<T>(y/n,x/n); }
bool operator<(const Point &other) const {if (x == other.x) return y < other.y;return x < other.x;}
Point<T> rotate(Point<T> center, ld angle) {
ld si = sin(angle * PI / 180.), co = cos(angle * PI / 180.);
ld y = this->y - center.y;
ld x = this->x - center.x;
return Point<T>(y * co - x * si + center.y, y * si + x * co + center.x);
}
ld distance(Point<T> other) {
T dy = abs(this->y - other.y);
T dx = abs(this->x - other.x);
return sqrt(dy * dy + dx * dx);
}
T norm() { return x * x + y * y; }
};
template<typename T>
struct Line {
Point<T> A, B;
Line(Point<T> A, Point<T> B) : A(A), B(B) {}
Line() {}
void input() {
A = Point<T>();
B = Point<T>();
A.input();
B.input();
}
T ccw(Point<T> &a, Point<T> &b, Point<T> &c) {
T res = a.x * b.y + b.x * c.y + c.x * a.y;
res -= (a.x * c.y + b.x * a.y + c.x * b.y);
return res;
}
bool isIntersect(Line<T> o) {
T p1p2 = ccw(A,B,o.A) * ccw(A,B,o.B);
T p3p4 = ccw(o.A,o.B,A) * ccw(o.A,o.B,B);
if (p1p2 == 0 && p3p4 == 0) {
pair<T,T> p1(A.y, A.x), p2(B.y,B.x), p3(o.A.y, o.A.x), p4(o.B.y, o.B.x);
if (p1 > p2) swap(p2, p1);
if (p3 > p4) swap(p3, p4);
return p3 <= p2 && p1 <= p4;
}
return p1p2 <= 0 && p3p4 <= 0;
}
pair<bool,Point<ld>> intersection(Line<T> o) {
if(!this->intersection(o)) return {false, {}};
ld det = 1. * (o.B.y-o.A.y)*(B.x-A.x) - 1.*(o.B.x-o.A.x)*(B.y-A.y);
ld t = ((o.B.x-o.A.x)*(A.y-o.A.y) - (o.B.y-o.A.y)*(A.x-o.A.x)) / det;
return {true, {A.y + 1. * t * (B.y - A.y), B.x + 1. * t * (B.x - A.x)}};
}
//@formula for : y = ax + b
//@return {a,b};
pair<ld, ld> formula() {
T y1 = A.y, y2 = B.y;
T x1 = A.x, x2 = B.x;
if(y1 == y2) return {1e9, 0};
if(x1 == x2) return {0, 1e9};
ld a = 1. * (y2 - y1) / (x2 - x1);
ld b = -x1 * a + y1;
return {a, b};
}
};
template<typename T>
struct Circle {
Point<T> center;
T radius;
Circle(T y, T x, T radius) : center(Point<T>(y,x)), radius(radius) {}
Circle(Point<T> center, T radius) : center(center), radius(radius) {}
Circle() {}
void input() {
center = Point<T>();
center.input();
cin>>radius;
}
bool circumference(Point<T> p) {
return (center.x - p.x) * (center.x - p.x) + (center.y - p.y) * (center.y - p.y) == radius * radius;
}
bool intersect(Circle<T> c) {
T d = (center.x - c.center.x) * (center.x - c.center.x) + (center.y - c.center.y) * (center.y - c.center.y);
return (radius - c.radius) * (radius - c.radius) <= d and d <= (radius + c.radius) * (radius + c.radius);
}
bool include(Circle<T> c) {
T d = (center.x - c.center.x) * (center.x - c.center.x) + (center.y - c.center.y) * (center.y - c.center.y);
return d <= radius * radius;
}
};
ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }
all3 __exgcd(ll x, ll y) { if(!y) return {x,1,0}; auto [g,x1,y1] = __exgcd(y, x % y); return {g, y1, x1 - (x/y) * y1}; }
ll __lcm(ll x, ll y) { return x / __gcd(x,y) * y; }
ll modpow(ll n, ll x, ll MOD = mod) { n%=MOD; if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }
bool solve(vvll A, vvll B) {
if(sz(A) == 1 or sz(A[0]) == 1) return A == B;
umll freq1, freq2;
rep(i,0,sz(A)) {
bool fl = i % 2;
rep(j,0,sz(A[0])) {
if(fl) {
freq1[A[i][j]] += 1;
freq1[B[i][j]] -= 1;
} else {
freq2[A[i][j]] += 1;
freq2[B[i][j]] -= 1;
}
fl = !fl;
}
}
for(auto& [_,v] : freq1) if(v) return false;
for(auto& [_,v] : freq2) if(v) return false;
return true;
}
int main() {
Code By Sumfi
cout.precision(12);
ll tc = 1;
cin>>tc;
rep(i,1,tc+1) {
ll n,m;
cin>>n>>m;
vvll A(n,vll(m)), B(n,vll(m));
readvvll(A);
readvvll(B);
if(solve(A,B)) pyes else pno
}
return 0;
}Angled Flip CodeChef Solution in C++14
#include <bits/stdc++.h>
#define int long long int
#define debug cout<<"K"
#define mod 1000000007
using namespace std;
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin>>t;
while(t--)
{
int n,m;
cin>>n>>m;
vector<vector<int>>a(n,vector <int>(m));
vector<vector<int>>b(n,vector <int>(m));
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
cin>>a[i][j];
}
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
cin>>b[i][j];
}
if(n==1||m==1)
{
if(a==b)
cout<<"YES\n";
else
cout<<"NO\n";
continue;
}
multiset<int>black1,black2,white1,white2;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if((i+j)%2==0)
{
black1.insert(a[i][j]);
black2.insert(b[i][j]);
}
else
{
white1.insert(a[i][j]);
white2.insert(b[i][j]);
}
}
}
if(black1==black2&&white1==white2)
cout<<"YES\n";
else
cout<<"NO\n";
}
return 0;
}Angled Flip CodeChef Solution in PYTH 3
# cook your dish here
for i in range(int(input())):
n,m = map(int,input().split())
a = []
b = []
for i in range(n):
a.append(list(map(int,input().split())))
for i in range(n):
b.append(list(map(int,input().split())))
if m==1 or n==1:
if a==b:
print("YES")
else:
print("NO")
else:
a1=[]
a2=[]
b1=[]
b2=[]
for i in range(n):
for j in range(m):
if (i+j)%2==0:
a1.append(a[i][j])
b1.append(b[i][j])
else:
a2.append(a[i][j])
b2.append(b[i][j])
a1.sort()
b1.sort()
a2.sort()
b2.sort()
if a1==b1 and a2==b2:
print("YES")
else:
print("NO")Angled Flip CodeChef Solution in C
#include <stdio.h>
#include <stdlib.h>
long long cmpfunc (const void * a, const void * b) {
return ( *(long long*)a - *(long long*)b );
}
int main(void) {
// your code goes herem
int t;
scanf("%d",&t);
l1:while (t--)
{
long long x ,y,a[100000000],b[100000000],c[100000000],d[100000000],i,j,e,f,k,l,m,n,l1=0,l2=0,l3=0,l4=0,flag1=1,flag2=1,flag3=1;
scanf("%lld%lld",&x,&y);
if(x==1||y==1)
{
k=0;
for(i=0;i<x;++i)
{
for(j=0;j<y;++j)
{
scanf("%lld",&a[k]);
k++;
}
}
l=0;
for(i=0;i<x;++i)
{
for(j=0;j<y;++j)
{
scanf("%lld",&b[l]);
l++;
}
}
for(i=0;i<(x*y);i++)
if(a[i]!=b[i])
flag3=0;
if(flag3)
printf("YES\n");
else
printf("NO\n");
goto l1;
}
k=0;
l=0;
for(i=0;i<x;++i)
{
for(j=0;j<y;++j)
{
if((i^j)%2)
{
scanf("%lld",&a[k]);
k++;
l1++;
}
else
{
scanf("%lld",&b[l]);
l++;
l2++;
}
}
}
m=0;
n=0;
for(i=0;i<x;++i)
{
for(j=0;j<y;++j)
{
if((i^j)%2)
{
scanf("%lld",&c[m]);
m++;
l3++;
}
else
{
scanf("%lld",&d[n]);
n++;
l4++;
}
}
}
qsort(a, l1, sizeof(long long), cmpfunc);
qsort(b, l2, sizeof(long long), cmpfunc);
qsort(c, l3, sizeof(long long), cmpfunc);
qsort(d, l4, sizeof(long long), cmpfunc);
if(l1!=l3||l2!=l4)
{
printf("NO\n");
goto l1;
}
else
{
for(i=0;i<l1;i++)
if(a[i]!=c[i])
flag1=0;
for(i=0;i<l2;i++)
if(b[i]!=d[i])
flag2=0;
if(flag1&&flag2)
printf("YES\n");
else
printf("NO\n");
}
}
return 0;
}Angled Flip CodeChef Solution in JAVA
/* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
static PrintWriter out=new PrintWriter((System.out));
public static void main(String args[])throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while(t-->0)
{
StringTokenizer str = new StringTokenizer(br.readLine());
int m = Integer.parseInt(str.nextToken());
int n = Integer.parseInt(str.nextToken());
HashMap<Integer,Integer> black=new HashMap<>();
HashMap<Integer,Integer> white=new HashMap<>();
for(int i=0; i<m; i++){
str = new StringTokenizer(br.readLine());
for(int j=0; j<n; j++){
int d = Math.abs((i-j)) % 2;
int num = Integer.parseInt(str.nextToken());
if(d==0){
black.put(num,black.getOrDefault(num,0) +1);
}else{
white.put(num,white.getOrDefault(num,0) +1);
}
}
}
for(int i=0; i<m; i++){
str = new StringTokenizer(br.readLine());
for(int j=0; j<n; j++){
int d = Math.abs((i-j)) % 2;
int num = Integer.parseInt(str.nextToken());
if(d==0){
black.put(num,black.getOrDefault(num,0) -1);
}else{
white.put(num,white.getOrDefault(num,0) -1);
}
}
}
if(n==1 || m==1){
out.println("NO");
continue;
}
boolean can = true;
for(Integer a: black.keySet()){
if(black.get(a)!=0){
can=false;
break;
}
}
if(!can){
out.println("NO");
continue;
}
for(Integer a: white.keySet()){
if(white.get(a)!=0){
can=false;
break;
}
}
if(!can){
out.println("NO");
}else{
out.println("YES");
}
}
out.close();
}
}Angled Flip CodeChef Solution in PYPY 3
tc = int(input())
for __ in range(tc):
n,m = map(int,input().split())
a=[]
b=[]
for i in range(n):
tmp = list(map(int,input().split()))
a.append(tmp)
for i in range(n):
tmp = list(map(int,input().split()))
b.append(tmp)
black1,black2,white1,white2=[],[],[],[]
if n==1 or m==1:
if a==b:
print("YES")
else:
print("NO")
continue
for i in range(n):
for j in range(m):
if (i+j)%2==0:
white1.append(a[i][j])
white2.append(b[i][j])
else:
black1.append(a[i][j])
black2.append(b[i][j])
white1.sort()
white2.sort()
black1.sort()
black2.sort()
if white1==white2 and black1==black2:
print("YES")
else:
print("NO")
Angled Flip CodeChef Solution in PYTH
class p3:
def __init__(self, n, m, a, b):
self.n = n
self.m = m
self.a = a
self.b = b
def solve(self):
if self.n == 1 or self.m == 1:
if self.a == self.b:
return 'YES'
else:
return 'NO'
a_odd = [0] * (self.n * self.m // 2)
b_odd = [0] * (self.n * self.m // 2)
if self.n * self.m % 2 == 1:
a_even = [0] * (len(a_odd) + 1)
b_even = [0] * (len(b_odd) + 1)
else:
a_even = [0] * len(a_odd)
b_even = [0] * len(b_odd)
po = 0
pe = 0
for i in range(0, self.n):
for j in range(0, self.m):
if (i + j) % 2 == 0:
a_even[pe] = self.a[i][j]
pe += 1
else:
a_odd[po] = self.a[i][j]
po += 1
po = 0
pe = 0
for i in range(0, self.n):
for j in range(0, self.m):
if (i + j) % 2 == 0:
b_even[pe] = self.b[i][j]
pe += 1
else:
b_odd[po] = self.b[i][j]
po += 1
a_odd.sort()
b_odd.sort()
a_even.sort()
b_even.sort()
'''
print a_odd
print b_odd
print a_even
print b_even
'''
if a_odd == b_odd and a_even == b_even:
return 'YES'
return 'NO'
def main():
t = int(raw_input())
while t > 0:
t -= 1
(n, m) = (int(x) for x in raw_input().split())
a = [[]] * n
for i in range(0, n):
a[i] = [ int(x) for x in raw_input().split()]
b = [[]] * n
for i in range(0, n):
b[i] = [ int(x) for x in raw_input().split()]
ob = p3(n, m, a, b)
print ob.solve()
if __name__ == '__main__':
main()Angled Flip CodeChef Solution in C#
using System;
public class Test
{
public static bool angledFlip(ref long[][] A, ref long[][] B, int N, int M) {
if (N == 1 || M == 1) return A == B;
long even = 0, odd = 0;
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
long tmp = A[i][j] ^ B[i][j];
if (((i + j) & 1) == 1) odd ^= tmp;
else even ^= tmp;
}
}
return even == 0 && odd == 0;
}
public static void Main()
{
// your code goes here
int T = int.Parse(Console.ReadLine());
while (T-- > 0) {
var num = Console.ReadLine().Split();
int N = int.Parse(num[0]);
int M = int.Parse(num[1]);
long[][] A = new long[N][];
for (int i = 0; i < N; i++) {
var arr = Console.ReadLine().Split();
A[i] = Array.ConvertAll(arr, long.Parse);
}
long[][] B = new long[N][];
for (int i = 0; i < N; i++) {
var arr = Console.ReadLine().Split();
B[i] = Array.ConvertAll(arr, long.Parse);
}
bool flag = angledFlip(ref A, ref B, N, M);
Console.WriteLine(flag ? "YES" : "NO");
}
}
}Angled Flip CodeChef Solution in GO
“package main
import (
"bufio"
"bytes"
"fmt"
"os"
"reflect"
)
func main() {
reader := bufio.NewReader(os.Stdin)
tc := readNum(reader)
var buf bytes.Buffer
for tc > 0 {
tc--
m, n := readTwoNums(reader)
A := make([][]int, m)
for i := 0; i < m; i++ {
A[i] = readNNums(reader, n)
}
B := make([][]int, m)
for i := 0; i < m; i++ {
B[i] = readNNums(reader, n)
}
res := solve(A, B)
if res {
buf.WriteString("YES\n")
} else {
buf.WriteString("NO\n")
}
}
fmt.Print(buf.String())
}
func readUint64(bytes []byte, from int, val *uint64) int {
i := from
var tmp uint64
for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
tmp = tmp*10 + uint64(bytes[i]-'0')
i++
}
*val = tmp
return i
}
func readInt(bytes []byte, from int, val *int) int {
i := from
sign := 1
if bytes[i] == '-' {
sign = -1
i++
}
tmp := 0
for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
tmp = tmp*10 + int(bytes[i]-'0')
i++
}
*val = tmp * sign
return i
}
func readString(reader *bufio.Reader) string {
s, _ := reader.ReadString('\n')
for i := 0; i < len(s); i++ {
if s[i] == '\n' || s[i] == '\r' {
return s[:i]
}
}
return s
}
func readNum(reader *bufio.Reader) (a int) {
bs, _ := reader.ReadBytes('\n')
readInt(bs, 0, &a)
return
}
func readTwoNums(reader *bufio.Reader) (a int, b int) {
res := readNNums(reader, 2)
a, b = res[0], res[1]
return
}
func readThreeNums(reader *bufio.Reader) (a int, b int, c int) {
res := readNNums(reader, 3)
a, b, c = res[0], res[1], res[2]
return
}
func readNNums(reader *bufio.Reader, n int) []int {
res := make([]int, n)
x := 0
bs, _ := reader.ReadBytes('\n')
for i := 0; i < n; i++ {
for x < len(bs) && (bs[x] < '0' || bs[x] > '9') && bs[x] != '-' {
x++
}
x = readInt(bs, x, &res[i])
}
return res
}
func solve(A, B [][]int) bool {
m := len(A)
n := len(A[0])
if m == 1 || n == 1 {
return reflect.DeepEqual(A, B)
}
// 奇偶性相同的位置可以随便变化
cnt := make([]map[int]int, 2)
for i := 0; i < 2; i++ {
cnt[i] = make(map[int]int)
}
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
x := A[i][j]
y := B[i][j]
k := (i + j) % 2
cnt[k][x]++
cnt[k][y]--
}
}
for _, cur := range cnt {
for _, v := range cur {
if v != 0 {
return false
}
}
}
return true
}Angled Flip CodeChef Solution Review:
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