prompt string | response string |
|---|---|
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | from sys import stdin,stdout
input=stdin.readline
import math,bisect
num = 102001
prime=[1]*num
prime[1]=0
prime[0]=0
for i in range(2,num):
j=i
while(j+i<num):
j+=i
prime[j]=0
l=[]
n,m=map(int,input().split())
for i in range(n):
t=list(map(int,input().split()))
l.append(t)
ans=60000000
for i in range(n):
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | from bisect import bisect_left as bisect
def primes(n):
correction = (n%6>1)
n = {0:n,1:n-1,2:n+4,3:n+3,4:n+2,5:n+1}[n%6]
sieve = [True] * (n/3)
sieve[0] = False
for i in xrange(int(n**0.5)/3+1):
if sieve[i]:
k=3*i+1|1
sieve[ ((k*k)/3) ::2*k]=[False]*((n/6-(k*k)/6-1)/... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;
import java.util.TreeMap;
public class B {
static final int MAX = 100500;
public static void main(String[] args) {
doIt();
}
static void doIt() {
Scanner sc = new Scanner(System.in);
PrintWriter pw = new PrintWriter(System.out);... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | def sieve(mx):
a = [0] * (mx + 1)
a[0] = a[1] = 1
for (i, e) in enumerate(a):
if e == 0:
for n in range(i*i, mx + 1, i):
a[n] = 1
return a
p = sieve(10**5 + 100)
for i in range(10**5 + 99, 0, -1):
p[i] *= p[i+1] +1
n,m = map(int, input().split())
cols = [0]*m
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main {
/**
* @param args
*/
public static void main(String[] args) {
boolean[] P = new boolean[1000000];
Arrays.fi... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.math.*;
import java.util.*;
import java.io.*;
import static java.lang.System.*;
public class Solution {
public static void main(String[] args) throws IOException
{
Scanner in= new Scanner(System.in);
int n,m;
int cnt=0;
int a[][] = new int[501][501];
int b[][] = new int[501][501];
int c[][] ... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | x,y=map(int,raw_input().split())
Maxprimos = 100100
primos=[True]*Maxprimos
primos[0]=False
primos[1]=False
marked={}
for i in range(2,Maxprimos):
if primos[i]:#Los vamos a marcar
for j in range(i+i,Maxprimos,i):
primos[j]=False
matriz=[]
for i in range(x):
matriz.append(map(int,raw_input().... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import math
n, m = input().split(" ")
line = int(n)
column = int(m)
limit = 100025
primes_list = [True for i in range(limit + 1)]
next_prime_distance = [0 for i in range(200000)]
def sieve_primes():
primes_list[0] = primes_list[1] = False
for i in range(2, int(math.sqrt(limit))):
if primes_list[i]:
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.io.*;
import java.math.*;
import java.util.*;
public class Task {
private static int[] primes;
private static int findNext(int a)
{
if(primes[primes.length/2] > a)
return findNext(a, 0, primes.length/2);
else if(primes[primes.length/2] < a)
return findNext(a, primes.length/2, primes.length-... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.util.*;
public class CF271B {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int[] prime = new int[1000000];
int size = 0;
prime[size++] = 2;
for (int i = 3; i <= 200000; i+=2)
if (isPrime(i))
prime[size++] = i;
int n = sc.nextInt(), m = sc.nextInt();
i... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #CRIVO
ncrivo = 1000000
crivo = [True for i in xrange(ncrivo)]
crivo[0] = crivo[1] = False
for i in xrange(2, ncrivo):
if not crivo[i]:
continue
for j in range(i * i, ncrivo, i):
crivo[j] = False
#lendo dados
n, m = map(int, raw_input().split())
data = []
for i in xrange(n):
data.append... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
int a[600][600];
bool isprime(int x) {
if (x == 1) return 0;
if (x == 2) return 1;
for (int i = 2; i * i <= x; i++) {
if (x % i == 0) return 0;
}
return 1;
}
int v[600], h[600];
bool pr[100007];
vector<int> prr;
int main() {
int n, m;
cin >> n >> m;
for ... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | def sieve(n):
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
a,m = map(int,input().split())
no=100100
prime = [True for i in range(no+1)]
sieve(no)
prime[0]=False
prime[1]=False
req=[0]*no
for i in range(no... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
long long k, p, h, b[555][555], a[555555];
long long mid, n, o, r, l;
bool y, m[1000001];
void s(long long r) {
for (long long i = 2; i * i <= r; i++) {
if (m[i] == false) {
for (long long j = i * 2; j <= r; j += i) {
m[j] = true;
}
}
}
for... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Stack;
import java.util.StringTokenizer;
public class Main {
static ArrayList<Integer> prime;
public static void sieve()
{
boolean[] isPrime = new bool... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.util.*;
import java.io.*;
public class B {
static int[] primes;
public static void genPrime(){
ArrayList<Integer> p = new ArrayList<>();
for(int i=2;i<=100000;i++){
boolean t = true;
for(int j=0;j<p.size();j++){
if(i%p.get(j)==0){
t = false;
break;
}
}
if(t)
p.a... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
const int N = 510;
const int M = 100100;
int prime[M + 10], a[N][N], b[N][N], n, m;
int main() {
prime[1] = 1;
for (int i = 2; i <= M; ++i)
if (prime[i] == 0)
for (int j = i + i; j <= M; j += i) prime[j] = 1;
cin >> n >> m;
for (int i = 1; i <= n; ++i)
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | def main():
limit = 110000
nums = [0]*limit
for i in xrange(2,limit):
if not nums[i]:
for j in xrange(i+i,limit,i):
nums[j]=1
delta = 0
for i in range(limit-1,0,-1):
if not nums[i]:
delta = 0
else:
nums[i]=delta
delta+=1
nums[1] = 1
A = []
B = []
C = []
nsums = []
msums = []
n,m = [in... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
template <typename T>
void _read(T& t);
template <typename T>
void _read(vector<T>& v);
template <typename T1, typename T2>
void _read(pair<T1, T2>& p);
template <typename T>
void _read(T& t) {
cin >> t;
}
template <typename T>
void _read(vector<T>& v) {
for (unsigned _... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.io.*;
import java.util.*;
public class B implements Runnable {
final int cnt = 200000;
private void Solution() throws IOException {
boolean[] prime = new boolean[2 * cnt + 1];
Arrays.fill(prime, true);
prime[0] = prime[1] = false;
for (int i = 2; i * i <= 2 * cnt; i++)
if (prime[i])
for... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | k=[0]*(200001)
primes=[]
for i in range(2,200001):
if k[i]==0:
primes.append(i)
for j in range(i,200001,i):k[j]=1
def bin(x):
lo,hi=0,len(primes)
ans=0
while lo<=hi:
mid = (hi+lo)//2
if primes[mid]==x:return 0
if primes[mid]<x:lo=mid+1
else:ans=mid;hi=mid-1
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | def to_prime(n):
step = 0
i = n
while not is_prime[i]:
step += 1
i += 1
return step
R = 10 ** 5 + 300
is_prime = [0] * (R + 1)
is_prime[1] = 1
d = 2
while d * d <= R:
if not is_prime[d]:
for i in range(d ** 2, R + 1, d):
is_prime[i] = 1
d += 1
for i in rang... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
int prime[200005];
void sieve() {
prime[0] = prime[1] = 1;
for (int i = 2; i < 200005; i += 1 + (1 & i)) {
if (!prime[i]) {
if (i <= 200005 / i)
for (int j = i * i; j < 200005; j += i) {
prime[j] = 1;
}
for (int j = i - 1; j >= ... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
bool primes[100005];
int sol[100005];
void sieve() {
memset(primes, true, sizeof primes);
primes[0] = primes[1] = false;
int i, j;
for (i = 2; i * i <= 100005; i++) {
if (primes[i]) {
for (j = i * i; j < 100005; j += i) primes[j] = false;
}
}
int a... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
bool compound[100030 + 1];
set<int> s;
unsigned long long int l, h;
int val, t;
void criba() {
for (int i = 2; i < 100030; i++) {
if (!compound[i]) {
s.insert(i);
val = i + i;
while (val <= 100030) {
compound[val] = true;
val += i;
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
vector<int> P;
char crivo[1000000];
int m[512][512];
int mm[512][512];
int N, M;
int main() {
int i, n, a;
memset(crivo, 0, sizeof(crivo));
for (i = 2; i < 1000000; i++) {
if (crivo[i]) continue;
P.push_back(i);
for (n = i + i; n < 1000000; n += i) crivo[n... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5, M = 6e2;
long long n, m, arr[M][M], cnt = 1e19, sum;
vector<long long> v;
bitset<N> isprime;
void sieve() {
isprime.set();
isprime[0] = isprime[1] = 0;
for (int i = 2; i <= N / i; i++) {
if (isprime[i]) {
for (int j = i * i; j < N; j +... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | def I(): return(list(map(int,input().split())))
def sieve(n):
p=2
primes=[True]*(n+1)
primes[1]=False
primes[0]=False
while(p*p<=n):
if primes[p]:
# // Update all multiples of p greater than or
# // equal to the square of it
# // numbers which are multiple of p and are
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | n=100100
p=[0,0]+[1]*(n)
p[0],p[1]=0,0
n1=int(n**0.5)
for i in range(2,n1):
if p[i]==1:
for j in range(i*i,n,i):
p[j]=0
for k in range(n,-1,-1):
if p[k]:
ind=k
p[k]=0
else:
p[k]=ind-k
lst=[]
x,y=map(int,input().split())
for j in range(x):
l=[]
for i in map(int,input().split()):
l.ap... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.io.*;
import java.util.*;
public class Round166Div2 {
public static void main(String[] args) throws IOException {
rd = new BufferedReader(new InputStreamReader(System.in));
pw = new PrintWriter(System.out);
st = new StringTokenizer(rd.readLine());
int N = Integer.parseInt(st.nextToken()), M = Inte... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | def Sieve():
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 1
prime[0]= False
prime[1]= False
n = 10**5 + 12345
prime = [True for i in range(n + 1)]
Sieve()
n, m = [int(j) for j in input().split()... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | def gen_prime(n):
num = [0 for i in range(n+1)]
num[2] = 1
num[3] = 1
for i in range(4, n+1):
isprime = True
for j in range(2, int(i**0.5)+1):
if num[j]:
if not i%j:
isprime = False
break
if isprime:
num[i] = 1
prime = []
for i in range(2, n+1):
if num[i]:
prime.append(i)
return ... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | I = lambda :map(int , raw_input().split())
n , m = I()
a = [I() for _ in xrange(n)]
N = 110000
p = [0] * N
_next = [0] * N
for i in xrange(2 , N):
if not p[i]:
for j in range(i * i , N , i):
p[j] = 1
cur_cnt = N - 1
for i in xrange(N - 1 , 1 , -1):
if not p[i]:
cur_cnt = i
_next... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import time
primes = list()
def generate_primes():
a=[0]*100010
for x in range(2,100004):
if a[x]==0:
primes.append(x)
j=2
while j*x<100004:
a[j*x]=1
j+=1
def bin_search(x):
#if x < primes[0]: return -1
left = 0
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | MaxN=110010
u=[0 for i in range(MaxN)]
u[0]=u[1]=1
for i in range(2,MaxN):
if u[i]:
continue
j=i*2
while j<MaxN:
u[j]=1
j+=i
i=MaxN-2
while i>0:
u[i]=u[i+1] if u[i] else i
i-=1
n,m=map(int,raw_input().split())
a=[map(int,raw_input().split()) for i in range(n)]
for i in range(... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.io.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
//solution classes here
public class Code {
//main solution here
static Scanner sc = new Scanner(System.in);
static PrintWriter out = new PrintWriter(System.out);
static long mod = 998244353;//(l... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | ncrivo = 1000000
crivo = [True for i in range(ncrivo)]
crivo[0] = crivo[1] = False
for i in range(2, ncrivo):
if crivo[i]:
for j in range(i * i, ncrivo, i):
crivo[j] = False
# lendo dados
n, m = map(int, input().split())
data = []
for i in range(n):
data.append(list(map(int, input().spl... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
void fast_in_out() {
std::ios_base::sync_with_stdio(NULL);
cin.tie(NULL);
cout.tie(NULL);
}
int dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0};
int dy[] = {-1, 1, 0, 0, -1, -1, 1, 1, 0};
int lx[] = {2, 2, -1, 1, -2, -2, -1, 1};
int ly[] = {-1, 1, 2, 2, 1, -1, -2, -2};
const lo... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
int a[501][501], p[100010];
int next(int n) {
int i;
for (i = n; p[i]; i++)
;
return i;
}
int main() {
int m, n, i, j, s, min = 10000000;
p[0] = p[1] = 1;
for (i = 2; i * i < 100010; i++)
if (p[i] == 0)
for (j = i * 2; j < 100010; j += i) p[j] = 1;
scanf("%d%d", &n, ... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
long long count(int n) {
if (n == 1)
return 0;
else if (n == 3)
return 8;
else
return count(n - 2) + (n * 4 - 4) * floor(n / 2.0);
}
bool isPrime(int n) {
int i, flag = 0;
if (n % 2 == 0)
flag = 1;
else {
for (i = 3; i <= sqrt(n); i += 2) {
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
template <typename T, typename U>
static inline void amin(T &x, U y) {
if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
if (x < y) x = y;
}
using namespace std;
const int N = 1e6;
int p[N];
void solve() {
long long int n, m, i, j, ans = INT_MAX;
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
bool v[110001];
int n, m, i, p[100001], c, j, Next[100001], k, w[501][501], S, Res = 999999999;
int main() {
p[c++] = 2;
for (i = 3; i < 110000; i += 2) {
if (v[i]) continue;
p[c++] = i;
for (j = 3 * i; j < 110000; j += i << 1) {
v[j] = true;
}
}
k = 0;
for (i = ... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | def main():
entrada = input().split(" ")
valor1 = int(entrada[0])
valor2 = int(entrada[1])
num1 = 100025
num2 = 2
matriz = []
primos = [True]*num1
primos = set_numeros_nao_primos(num1, num2, primos)
for i in range(valor1):
linha = list(map(int,input().split()))
matri... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class BOOL {
private static FastReader in =new FastReader();
static ArrayList<Integer> A=new ArrayList();
static HashMap <Integer,Integer> Map =new HashMap();
static boolean B[][];
sta... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
bool isPrime(int x) {
if (x < 2) return false;
for (int i = 2; i <= sqrt(x); i++)
if (!(x % i)) return false;
return true;
}
int main() {
int nextPrime[100000];
for (int i = 0; i < 100000; i++) {
for (int j = i + 1;; j++)
if (isPrime(j)) {
ne... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import math
def prime(n):
ok= True
for i in range(2, int(math.sqrt(n))):
if(n%i==0):
ok = False
break
if(ok):
return True
else:
return False
def fact(a,b):
ans = 1
for i in range(a, b+1):
ans*= i
return str(ans)-1
def comb(n, c):
re... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | num1= 102001
num2= 2
prime=[1]*num1
prime[1]=0
prime[0]=0
for i in range(num2,num1):
j=i
while(j+i<num1):
j+=i
prime[j]=0
l=[]
n,m=map(int,input().split())
for i in range(n):
t=list(map(int,input().split()))
l.append(t)
ans=100000
for i in range(n):
tot=0
for j in range(m):
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.util.*;
import java.io.*;
public class palin {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
boolean prime[] = new boolean[100004];
for (int i = 2; i < 50003; i++) {
for (int j = 2; i * j < 100004; j++) {
prime... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
const int N = 1000001;
bool crib[N];
void criba() {
memset(crib, true, sizeof(crib));
crib[1] = false;
for (int i = 2; i <= N / 2; i++) {
if (crib[i] == true) {
for (int j = 2 * i; j <= N; j += i) {
crib[j] = false;
}
}
}
}
int mov_fila(i... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.util.LinkedList;
import java.util.Scanner;
public class Contest_5C {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String[] input1 = sc.nextLine().split(" ");
int rows = Integer.parseInt(input1[0]);
int cols = Integer.parseInt(input1[1]);
String[][] input2 = n... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | # JOAO MARCELO - 2021
# 271B
import math
from bisect import bisect_left
maxi = 100010
n, m = [int(x) for x in input().split()]
tag = [0] * maxi
lines = [0] * n
cols = [0] * m
prime = [0] * maxi
tag[0] = tag[1] = 1
count = 0
for i in range(2, maxi):
if(not tag[i]):
prime[count] = i
count += 1
j ... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.util.NavigableSet;
import java.util.Scanner;
import java.util.TreeSet;
public class B271 {
static NavigableSet<Integer> primes = new TreeSet<>();
static void initPrimes(int MAXN) {
int sqrtN = (int)Math.sqrt(MAXN);
boolean[] composite = new boolean[MAXN];
composite[0] = c... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... |
import java.util.*;
public class B271 {
public static boolean[] getListOfPrimalNumber() {
int n = 100004;
boolean[] arr = new boolean[n];
for (int p = 2; p < n; p++) arr[p] = true;
for (int p = 2; p < n; p++)
if (arr[p]) {
for (int j = 2 * p; j < n; j +... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | m = 1000000
p=2
d = [-1]*(m)
while p**2<=m:
if d[p]==-1:
for i in range(p*2, m , p):
d[i]=0
p+=1
d[0]=0
d[1]=0
w=[]
#print(d[:10])
g=0
for j in range(m-1,1,-1):
if d[j]!= -1:
w.append(g)
else:
g=j
w.append(g)
w.append(2)
w.append(2)
sq=w[::-1]
#print(... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import math
import sys
def get_primes_distances(limit):
distances = [0] * (limit +1)
sieve = [True] * (limit + 1)
sieve[0] = False
sieve[1] = False
i = 2
while (i * i <= limit):
if (sieve[i] == True):
for j in range(i * 2, (limit+1), i):
sieve[j] = False
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | def arr_inp(n):
if n == 1:
return [int(x) for x in stdin.readline().split()]
elif n == 2:
return [float(x) for x in stdin.readline().split()]
else:
return [str(x) for x in stdin.readline().split()]
def count_prime(n):
prim = defaultdict(lambda: 1, {i: 1 for i in range(n + 1)})
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class B {
private static final String Object = null;
static BufferedReader in;
static StringTokenizer st;
static Pri... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
const int N = 1000000;
int prime[N], primes[N];
void sieve() {
for (int i = 2; i <= N; i++) prime[i] = 1;
for (int i = 2; i * i <= N; i++)
if (prime[i])
for (int y = i * i; y <= N; y += i) prime[y] = 0;
}
int main() {
sieve();
int idx = 0;
for (int i = 0... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | def buil_prime_dict(n):
a =[x for x in range(n+1)]
a[1] = 0
lst = []
i = 2
while i <= n:
if a[i] != 0:
lst.append(a[i])
for j in range(i, n+1, i):
a[j] = 0
i += 1
return lst
prime_dict=buil_prime_dict(10**5+100)
def diff_search_n... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
;
long long n, m;
cin >> n >> m;
long long mat[n][m];
for (long long i = 0; i < (long long)n; i++) {
for (long long j = 0; j < (long long)m; j++) {
cin >> mat[i][j];
}
}
vector<lon... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.util.Scanner;
public class P271B
{
public static void main (String[] args)
{
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int m = in.nextInt();
int[][] matrix = new int[n][m];
boolean[] primes = Eratesten(100003);
// int arrayMax = 1;
// int index = 0;
for (int i=0; i<n; i++)... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
int n, m;
int ers[101000];
int mp[101000];
int a[510][510];
int ans;
int main() {
int i, j, tmp;
ers[0] = 1;
ers[1] = 1;
for (i = 2; i < 100100; i++) {
if (ers[i] == 0) {
for (j = i * 2; j < 100100; j += i) {
ers[j] = 1;
}
}
}
j = 999999;
for (i = 10010... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Arrays;
import java.util.Enumeration;
import java.util.Iterator;
impo... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | def _min(x, y):
if x < y:
return x
if y <= x:
return y
n = 100100
prime = [True for i in range(n+1)]
prime[0], prime[1] = False, False
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
n = 100100
nextPrime = [0 for _ in range(n+1)]
curren... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-6;
const int PRECISION = 20;
const int MOD = 1e9 + 7;
struct node {
long long val;
vector<long long> formula;
node() { val = -1; }
};
struct group {
long long mul, last, gcd;
group(long long m, long long l, long long lm) {
mul = m;
la... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | # ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
long long n, m, ans = 1e18;
long long mat[505][505];
bool OK[100005];
vector<long long> primes;
void Sieve() {
for (long long i = 3; i < 100005; i += 2) OK[i] = 1;
for (long long i = 3; i < 100005; i += 2)
if (OK[i])
for (long long j = i * i; j < 100005; j += ... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | n,m = map(int,raw_input().split())
arr = []
for i in range(n):
arr.append(list(map(int,raw_input().split())))
primes = []
prime = [0 for x in range(100100)]
prime[0] = prime[1] = 1
for i in range(2,100100):
if not prime[i]:
primes.append(i)
for j in range(i+i,100100,i):
prime[j] = 1
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.util.Scanner;
public class PrimeMatrix {
private static Scanner read = new Scanner(System.in);
public static boolean isPrime(int a) {
if (a < 2)
return false;
if (a != 2 && a % 2 == 0)
return false;
for (int i = 3; i * i <= a; i = i + 2) {
if (a % i == 0)
return false;
}
return t... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... |
n = 100005
primes = [True]*(n+1)
primes[0] = False
primes[1] = False
p = 2
while(p*p<=n):
if primes[p]:
for i in range(p*p,n+1,p):
primes[i] = False
p += 1
plist = []
for i,val in enumerate(primes):
if val:
plist.append(i)
def closestPrime(a):
if primes[a]:
return(a)
first = 0
last = len(plist) - 1... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import sys
import math as mt
import bisect as bi
import collections as cc
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
N = 10**5+100
prime = []
pr = [0]*(N)
pr[0] = pr[1] = 1
for i in range(2,N):
if not pr[i]:
prime.append(i)
for j in range(2*i,N,i):
pr[j] = 1
n... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | //package com.example.programming;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import ... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
int a[505][505];
bool isprime[100005];
int prime[100005], x = 0;
int bse(int temp) {
int beg = 0, last = x - 1, an = 10000000000;
while (beg <= last) {
int mid = beg + (last - beg) / 2;
if (prime[mid] == temp)
return prime[mid];
else if (prime[mid] > t... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.io.*;
import java.util.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
public class PrimeMatrix {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
//READ----------------------------------------------------
int n = sc.nextIn... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.util.*;
public class PrimeMatrix {
static boolean[] isPrime;
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int min = Integer.MAX_VALUE;
flagPrimes(999999);
int[][] arr... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | /*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
import java.io.*;
import java.math.BigInteger;
import java.util.*;
import java.text.*;
public class cf271b {
static BufferedReader br;
static Scanner sc;
static PrintWriter out;
public static void ini... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.util.*;
public class Main {
// n levels and 4 vertex
public static void main(String[] args){
Scanner sc= new Scanner(System.in);
ArrayList<Integer> p= new ArrayList<>();
for (int i = 2; i <= 100000+10; i++) {
boolean prime=true;
for (int j= 2; j <= Math.sqrt(i); j++) {
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.util.*;
import java.io.*;
import javafx.util.Pair;
import java.math.BigInteger;
import java.text.*;
public class cf2 {
static long mod = (long)1e9 + 7;
static long mod1 = 998244353;
static FastScanner f;
static PrintWriter pw = new PrintWriter(System.out);
static BufferedReader br;
s... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1}, dy[] = {1, -1, 0, 0, 1, -1, -1, 1};
template <class T>
void cmin(T& a, T b) {
if (b < a) a = b;
}
template <class T>
void cmax(T& a, T b) {
if (b > a) a = b;
}
template <class T>
int len(const T& c) {
return (int)c.size();
}
tem... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | LIMIT = 100000+4
l, c = map(int, raw_input().split())
m = [ list(map(int, raw_input().split())) for x in xrange(l)]
def is_prime(num):
if num == 2 or num == 3: return True
if num < 2 or num%2 == 0: return False
if num < 9: return True
if num%3 == 0: return False
r = int(num**0.5)
f = 5
while f <= r:
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.util.Scanner;
public class PrimeMatrix {
public static void main(String[] args) {
System.out.println();
Scanner sc = new Scanner(System.in);
boolean prime[] = SOE(1000000);
int n = sc.nextInt();
int m = sc.nextInt();
int matrix[][] = new int[n][m];
int row[][] = new int[n][m];
int col... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.util.Scanner;
public class PrimeMatrix {
private static int MAX_INPUT = 100003;
private static boolean[] generateAllPrimes() {
boolean primeNumbers[] = new boolean[MAX_INPUT + 1];
for (int i = 0; i <= MAX_INPUT; i++) {
primeNumbers[i] = true;
}
for (i... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | n, m = [int(s) for s in raw_input().split()]
A = [None] * n
for i in range(n):
row = [int(s) for s in raw_input().split()]
A[i] = row
Amax = max([i for a in A for i in a])
def generate_primes(Amax):
Amax = max(Amax, 10)
numbers = [i for i in range(100004)]
mapping = {n: True for n in numbers}
mapping[0]... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import bisect
import collections
import copy
import functools
import heapq
import itertools
import math
import random
import re
import sys
import time
import string
from typing import List
sys.setrecursionlimit(99999)
p = []
mx = 10**5+1000
f = [0]*mx
for i in range(2,mx):
if f[i]==0:
p.append(i)
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Comparator;
public class practiceQuestions1 {
static class Reader {
final private int BUFFER_SIZE = 1 << 12;
boolean consume = false;
private byte[... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
int a[1000000], n, m, tmp, x, row[501], col[501];
vector<int> primes;
void generate_primes() {
for (long long i = 2; i < 1000000; ++i) {
if (!a[i]) {
primes.push_back(i);
for (long long j = i * i; j < 1000000; j += i) {
a[j] = 1;
}
}
}
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | from sys import stdin,stdout,setrecursionlimit
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush,nlargest
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm , accumulate
from bis... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import os
import sys
import math
import heapq
from decimal import *
from io import BytesIO, IOBase
from collections import defaultdict, deque
def r():
return int(input())
def rm():
return map(int,input().split())
def rl():
return list(map(int,input().split()))
def prime(p):
if p==1:
return 0
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
long long n, m, kompos[1100010], a[1100][1100], b[1100][1100], temp;
int main() {
cin >> n >> m;
kompos[1] = 1;
for (long long i = 2; i * i <= 1100000; i++) {
if (!kompos[i]) {
long long j = i * i;
while (j <= 1100000) {
kompos[j] = 1;
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
bool notprime[100101];
void isprime() {
int i, j;
notprime[1] = true;
for (i = 2; i <= 100100; i++) {
if (notprime[i] == false) {
for (j = i + i; j <= 100100; j += i) {
notprime[j] = true;
}
}
}
}
int cnt[501][501];
int main() {
int n, ... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.io.*;
import java.util.*;
public class Main {
static boolean[] isPrime;
static ArrayList<Integer> primes;
public static void sieve(){
isPrime = new boolean[1000_051];
primes = new ArrayList<>();
Arrays.fill(isPrime, true); isPrime[1] = isPrime[0] = false;
for (... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
const int oo = INT_MAX;
int arr[1000000];
vector<int> p;
void sieve() {
arr[0] = arr[1] = 1;
for (int i = 2; i < 1000000; ++i) {
if (!arr[i]) {
p.push_back(i);
for (long long j = i; j * i < 1000000; ++j) {
arr[i * j] = 1;
}
}
}
}
int ... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | maxn = 100100
ar = [1 for i in range(maxn)]
ar[0], ar[1] = 0, 0
for i in range(2, maxn):
if ar[i]:
for j in range(i, (maxn - 1) // i + 1):
ar[i * j] = 0
dst = maxn
d = [dst for i in range(maxn)]
for i in reversed(range(maxn)):
if ar[i]: dst = 0
d[i] = min(d[i], dst)
dst += 1
n, m = map(int, input(... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import bisect
def seive():
seive_arr = [True]*1000100
seive_arr[0] = False
seive_arr[1] = False
for i in range(2, 1000002):
if(seive_arr[i]):
for j in range(i*i, 1000002, i):
seive_arr[j] = False
primes = []
for i in range(1000002):
if(seive_arr[i]):
primes.append(i)
return primes
primes = seive(... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
public class Main {
private static MyScanner in;
private static PrintStream out;
public static void main(String[] args) throws IOException {
boolean LOCAL_TEST = false;
out = System.out;
if (LOCAL_TE... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import static java.lang.System.*;
import static java.lang.Math.*;
import java.math.*;
import java.util.*;
import java.io.*;
public class CodeForces {
public static void main(String[] args) {
boolean isPrime[]=new boolean[1000000];
isPrime[0]=true;
for(int i=1; i<1000000; i++){
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
vector<int> sv;
bool mrk[300000];
int grid[700][700];
void filla() {
for (int i = 2; i <= 200000; i++) {
if (!mrk[i]) {
sv.push_back(i);
for (int j = 2; j * i <= 200000; j++) mrk[j * i] = 1;
}
}
}
int main() {
filla();
int n, m;
cin >> n >> m;
... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... |
import java.io.*;
import java.util.*;
public class CN167B {
static FastScanner in =new FastScanner(System.in);
static boolean isComposite[] = new boolean[1000001];
static int INF = Integer.MAX_VALUE/2;
public static void main (String args[]){
GharbalErnesten(1000000);
isComposite[1]=true;
isComposite[0]=tru... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | #include <bits/stdc++.h>
using namespace std;
struct hash_pair {
template <class T1, class T2>
size_t operator()(const pair<T1, T2>& p) const {
auto hash1 = hash<T1>{}(p.first);
auto hash2 = hash<T2>{}(p.second);
return hash1 ^ hash2;
}
};
long long t;
vector<long long> v;
long long arr[1000005];
void... |
Problem: You've got an n Γ m matrix. The matrix consists of integers. In one move, you can apply a single transformation to the matrix: choose an arbitrary element of the matrix and increase it by 1. Each element can be increased an arbitrary number of times.
You are really curious about prime numbers. Let us remind y... | import bisect
n, m = map(int, raw_input().split())
matrix = []
for i in xrange(n):
matrix.append(list(map(int, raw_input().split())))
primes = [i for i in xrange(101000)]
primes[0] = -1
primes[1] = -1
only_primes = []
for i in xrange(len(primes)):
if primes[i] != -1:
only_primes.append(primes[i])
... |
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