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Prime Pattern CodeChef Solution
Problem -Prime Pattern 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.
Prime Pattern CodeChef Solution in C++14
“#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 = 202020;
const ll mod = 1e9 + 7;
// 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; }
const static int bits = 19, mask = (1<<19) - 1;
unsigned long long mul(unsigned long long x, unsigned long long y, unsigned long long M)
{
if (x <= UINT_MAX && y <= UINT_MAX) return x * y % M;
unsigned long long rslt = y * (x & mask) % M;
while (x >>= bits) rslt = (rslt + (y = (y << bits) % M) * (x & mask)) % M;
return rslt;
}
static int test(unsigned long long n, unsigned long long a) {
unsigned long long s, t, nmin1 = n - 1;
int r;
for (s = nmin1, r = 0; !(s & 1); s >>= 1, r++) ;
for(t = a; s >>= 1; )
{
a = mul(a, a, n);
if (s & 1) t = mul(t, a, n);
}
if (t == 1 || t == nmin1) return 1;
while (--r) if ((t = mul(t, t, n)) == nmin1) return 1;
return 0;
}
static int millerRabin(unsigned long long n) {
if(n<29) return n==2 || n==3 || n==5 || n==7 || n==11 || n==13 || n==17 || n==19 || n==23;
if(!(n&1 && n%3 && n%5 && n%7 && n%11 && n%13 && n%17 && n%19 && n%23)) return 0;
return test(n, 2) && test(n, 1215) &&(n==17431 || test(n, 34862) && (n==3281359 || test(n, 574237825)));
}
ll work(ll x, ll y) {
ll ma = max(abs(x), abs(y));
if(x == ma) return 4 * ma * ma - 4 * ma + 1 + y - (1 - ma);
if(y == ma) return 4 * ma * ma - 2 * ma + abs(x - ma);
if(x == -ma) return 4 * ma * ma + abs(y - ma);
if(y == -ma) return 4 * ma * ma + 2 * ma + abs(x + ma);
return -1;
}
ll solve(ll a, ll b) {
if(millerRabin(work(a,b))) return 0;
rep(res,1,INF) {
rep(x,1,res + 1) if(millerRabin(work(a + x, b + res - x))) return res;
rep(x,1-res,1) if(millerRabin(work(a + x, b + res + x))) return res;
rep(x,-res,0) if(millerRabin(work(a + x, b - res - x))) return res;
rep(x,0,res) if(millerRabin(work(a + x, b - res + x))) return res;
}
return -1;
}
int main() {
Code By Sumfi
cout.precision(12);
ll tc = 1;
cin>>tc;
rep(i,1,tc+1) {
ll a,b;
cin>>a>>b;
print(solve(a,b));
}
return 0;
}Prime Pattern CodeChef Solution in C
“#include <stdio.h>
#include <math.h>
const static int bits = 19, mask = (1<<19) - 1;
#define UINT_MAX 32767
#define NN 4000001
#define RR 2000000
long long cor2ind(long long x,long long y)
{
long long sum=0;
if(x-y>0 && x+y<=1) // south
{
sum=(y-1)*(4*y+2)+2;
sum+=x-y;
}
else if(x+y>1 && x-y>=0) // east
{
sum=(4*x-6)*x+2;
sum+=2*x-1;
sum+=x+y-1;
}
else if(x-y<0 && x+y>=0) // north
{
sum=(4*y-6)*y+2;
sum+=(y-1)*4+2;
sum+=y-x;
}
else if(x+y<0 &&x-y<=0) // west
{
sum=(8*x+12)*x/2+2;
sum+=(-x-1)*6+4;
sum+=-x-y;
}
return sum;
}
unsigned long long mul(unsigned long long x, unsigned long long y, unsigned long long M)
{
if (x <= UINT_MAX && y <= UINT_MAX) return x * y % M;
unsigned long long rslt = y * (x & mask) % M;
while (x >>= bits) rslt = (rslt + (y = (y << bits) % M) * (x & mask)) % M;
return rslt;
}
static int test(unsigned long long n, unsigned long long a) {
unsigned long long s, t, nmin1 = n - 1;
int r;
for (s = nmin1, r = 0; !(s & 1); s >>= 1, r++) ;
for(t = a; s >>= 1; )
{
a = mul(a, a, n);
if (s & 1) t = mul(t, a, n);
}
if (t == 1 || t == nmin1) return 1;
while (--r) if ((t = mul(t, t, n)) == nmin1) return 1;
return 0;
}
static int isprime(unsigned long long n) {
if(n<29) return n==2 || n==3 || n==5 || n==7 || n==11 || n==13 || n==17 || n==19 || n==23;
if(!(n&1 && n%3 && n%5 && n%7 && n%11 && n%13 && n%17 && n%19 && n%23)) return 0;
return test(n, 2) && test(n, 1215) &&(n==17431 || test(n, 34862) && (n==3281359 || test(n, 574237825)));
}
int main()
{
long long T, x0, y0, dist;
long long found, x, y;
scanf("%lld",&T);
while(T--)
{
scanf("%lld%lld",&x0,&y0);
found=0;
if(isprime(cor2ind(x0,y0)))
dist=0;
else
for(dist=1;;dist++)
{
// here x and y are relative coordinates
for(x=1;x<=dist;x++) // first quadrant
{
y=dist-x;
if(isprime(cor2ind(x+x0,y+y0)))
{
found=1;
break;
}
}
if(found) break;
for(x=1-dist;x<=0;x++) // second quadrant
{
y=dist+x;
if(isprime(cor2ind(x+x0,y+y0)))
{
found=1;
break;
}
}
if(found) break;
for(x=-dist;x<0;x++) // third quadrant
{
y=-dist-x;
if(isprime(cor2ind(x+x0,y+y0)))
{
found=1;
break;
}
}
if(found) break;
for(x=0;x<dist;x++) // fourth quadrant
{
y=-dist+x;
if(isprime(cor2ind(x+x0,y+y0)))
{
found=1;
break;
}
}
if(found) break;
}
printf("%lld\n",dist);
}
return 0;
} Prime Pattern CodeChef Solution in JAVA
“
import java.math.BigInteger;
import java.util.Scanner;
class PrimeNo {
public static int mod(int x){
return Math.abs(x);
}
public static boolean isPrime(long n){
BigInteger b = new BigInteger(Long.toString(n));
return b.isProbablePrime(10);
}
public static long index(int x1,int y1) {
// System.out.print(x1+" "+y1+" ");
int c = Math.max(mod(x1), mod(y1));
long y=(long)y1;
long x= (long)x1;
long indexSquare=0;
if (c == Math.abs(y1)) {
if (y1 > 0) {
indexSquare = 4 *y*y-y - x;
} else {
indexSquare = 4*y*y-3*y + x;
}
}else{
if (x1 >0) {
indexSquare = 4*x*x-3*x+y;
} else {
indexSquare = 4*x*x-(x)-y;
}
}
// System.out.println(indexSquare);
return indexSquare;
}
public static int distance(int x,int y) {
for (int k = 0; ; k++) {
int i = k;
// System.out.println(i);
for (int j = 0;j<=k; j++,i--) {
// System.out.println(j);
if(isPrime(index(x+i, y+j))||isPrime(index(x-i, y-j))||isPrime(index(x-j,y+i))||isPrime(index(x+j, y-i))){
return k;
}
}
}
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int T = scanner.nextInt();
for (int t = 0; t < T; t++) {
int x = scanner.nextInt();
int y = scanner.nextInt();
System.out.println(distance(x,y));
}
}
}
Prime Pattern CodeChef Solution in PYTH
“def millerRabin(n):
# Returns true if probable prime, false if composite
bases = [2,3,5,7,11,13,17,19,23]
nm1 = n-1
m = nm1
d = 0
if n in bases:
return True
if n < 2:
return False
while not m&1:
d += 1
m >>= 1
for a in bases:
done_for_base = False
b = pow(a,m,n)
if b == 1 or b == nm1 :
#done_for_base = True
continue
for k in range(d-1):
b = pow(b,2,n)
if b == 1:
return False
elif b == nm1:
done_for_base = True
break
if not done_for_base:
return False
return True
def valAt(x,y):
if x>y:
if x+y > 0:
n = 4*x*x - 3*x + y
else :
n = 4*y*y - 3*y + x
else:
if x+y > 0:
n = 4*y*y - x - y
else :
n = 4*x*x - y - x
return n
def solveCase(x,y):
r = 0
n = valAt(x,y)
if millerRabin(n):
return r
while True:
r += 1
for k in range(r):
n1 = valAt(x + r-k, y + 0-k)
n2 = valAt(x + 0-k, y - r+k)
n3 = valAt(x - r+k, y + 0+k)
n4 = valAt(x + 0+k, y + r-k)
for n in [n1,n2,n3,n4]:
if millerRabin(n):
return r
def main():
T = int(raw_input(''))
for t in range(T):
x,y = [ int(i) for i in raw_input().split() ]
ans = solveCase(x,y)
print ans
main()Prime Pattern CodeChef Solution Review:
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