prompt string | response string |
|---|---|
Problem: You've got an n × 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... | # specify list capacity
MaxN = 101002
prime_list = [0]*MaxN
prime_list[0] = prime_list[1] = 1
# mark all composite number with 1
for i in range(2, MaxN):
if prime_list[i] == 1:
continue
j = i*2
while j < MaxN:
prime_list[j] = 1
j += i
# then replace all '0' and '1' with prime number
i = MaxN-2
while i > 0:... |
Problem: You've got an n × 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 generate_primes(n):
# Create a boolean array "prime[0..n]" and initialize
# all entries it as true. A value in prime[i] will
# finally be false if i is Not a prime, else true.
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
# If prime[p] is not changed, then it is a p... |
Problem: You've got an n × 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 int binarySearch(int valor, int[] datos) {
int left=0,
right=datos.length-1,
avg;
while (left<=right) {
avg=(right+left)/2;
if(datos[avg]==valor) {
return avg;
}else if(datos[avg]<valor && valor<datos[avg+1]) {
avg++;
... |
Problem: You've got an n × 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.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
public class B {
static StringTokenizer st;
static Buffered... |
Problem: You've got an n × 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,math
isprime = [0]*1000010
all_prime = [2]
next_prime = [0]*1000010
def seive():
isprime[1] = isprime[0] = 1
limit = int(math.sqrt(1000010))+2
for i in range(4,1000010,2):
isprime[i] = 1
for i in range(3,1000010,2):
if(not isprime[i]):
all_prime.append(i)
... |
Problem: You've got an n × 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... | # Problem: B. Prime Matrix
# Contest: Codeforces - Codeforces Round #166 (Div. 2)
# URL: https://codeforces.com/problemset/problem/271/B
# Memory Limit: 256 MB
# Time Limit: 2000 ms
#
# Powered by CP Editor (https://cpeditor.org)
from sys import stdin, stdout
def INI():
return int(stdin.readline())
def INL():
r... |
Problem: You've got an n × 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 = map(int, input().rstrip().split())
# dp = [[0] * m for _ in range(n)]
row = [0] * n
col = [0] * m
# mat = []
def isPrime(x):
if x in [2, 3, 5, 7, 11, 13, 17, 19]:
return True
if x == 1:
return False
for i in range(2,int( math.sqrt(x)) + 2):
if x % i == 0:
... |
Problem: You've got an n × 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... | numMax = 102001
numMin = 2
prime = [1] * numMax
prime[1] = 0
prime[0] = 0
for i in range(numMin, numMax):
j = i
while(j+i < numMax):
j += i
prime[j] = 0
linha, coluna = map(int, raw_input().split())
matrix = []
for i in range(linha):
values = list(map(int, raw_input().split()))
matrix.append(values)
saida ... |
Problem: You've got an n × 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;
string uppercase(string s) {
transform(s.begin(), s.end(), s.begin(), ::toupper);
return s;
}
string lowercase(string s) {
transform(s.begin(), s.end(), s.begin(), ::tolower);
return s;
}
set<pair<int, pair<string, int>>> sp;
vector<vector<int>> v2d(5, vector<int>(5... |
Problem: You've got an n × 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.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
public class CF_271_B_PRIME_MATRIX {
static final int MAX = (int) (10e6+1);
static boolean notPrime [] = new ... |
Problem: You've got an n × 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... | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Mon Apr 20 20:37:49 2020
@author: narayanaaramamurthy
"""
a,b=map(int,input().split())
c=100030
f=[0]*c
f[1]=1
for i in range(2,c):
if f[i]==0:
for j in range(i+i,c,i):
f[j]=1
t=0
for i in range(c-1,0,-1):
if f[i]==0:
... |
Problem: You've got an n × 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 collections
import bisect
prime = [-1] * 200000
i = 2
prime_list = []
while 200000 > i:
if prime[i - 1] == -1:
prime[i - 1] = 0
temp = i * 2
prime_list.append(i)
while temp < 200000:
prime[temp - 1] = 1
temp += i
if i == 2:
i += 1
else:
i += 2
n, m = map(int, sys.stdin.readlin... |
Problem: You've got an n × 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<long long> prime;
bool isprime[1000001];
void find_prime() {
memset(isprime, true, sizeof isprime);
for (long long i = 2; i <= sqrt(1000000); i++) {
if (isprime[i] == true) {
for (long long j = i * i; j <= 1000000; j += i) {
isprime[j] = false;
... |
Problem: You've got an n × 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
R = lambda: map(int, raw_input().split())
n, m = R()
a = [R() for i in range(n)]
M = 110000
p = [0, 0] + [1] * M
for i in range(2, M):
for j in xrange(2, int(math.sqrt(i) + 1)):
if i % j == 0:
p[i] = 0
break
last = 10**10
for i in reversed(range(1, M)):
if p[i]... |
Problem: You've got an n × 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 = map(int,input().split())
grid = [list(map(int, input().split())) for _ in range(n)]
def seive():
s = 10**5 + 10
primes = [True] * (s)
primes[0] = False
primes[1] = False
for i in range(2,int(math.sqrt(s))+1):
j = 2
while(j*i<s):
#print(j*i)
... |
Problem: You've got an n × 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 LIMIT = 1000005;
int sieve[LIMIT + 1];
int primes[LIMIT + 1];
int mark_primes() {
int primeCount = 1;
for (int i = 0; i <= LIMIT; ++i) sieve[i] = 0;
for (int i = 2; i <= LIMIT; ++i) {
if (!sieve[i]) {
primes[primeCount] = i;
sieve[i] = primeC... |
Problem: You've got an n × 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 codeforces1;
import java.util.*;
import java.util.function.Function;
import java.util.stream.Collectors;
import java.io.*;
import java.math.*;
import java.text.*;
public class B166 {
static InputReader in = new InputReader(System.in);
static OutputWriter out = new OutputWriter(System.out);
public stat... |
Problem: You've got an n × 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 second
{
static long fast_power(long a,long n,long m)
{
if(n==1)
{
return a%m;
}
if(n%2==1)
{
long power = fast_power(a,(n-1)/2,m)%m;
return ((a%m) * ((power*power)%m))%m;
}
long power = fast_power(a,n/2,m)%m;
return (power*power)%m;
... |
Problem: You've got an n × 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.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
public class Task271B {
public static void main(String... args) throws NumberFormatException,
IOException {
Solutio... |
Problem: You've got an n × 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 × 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 Main2 {
public static void main(String args[]){
Scanner input = new Scanner(System.in);
int row = input.nextInt();
int col = input.nextInt();
int[][] matrix = new int[row][col];
for(int i = 0 ; i < row ; i++){
for(int j = 0 ;... |
Problem: You've got an n × 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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
im... |
Problem: You've got an n × 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<long long int> prime;
bool arr[100009];
void sieve() {
long long int k = sqrt(100009);
for (int i = 3; i <= k; i += 2) {
if (arr[i] == 0) {
for (long long int j = i * i; j < 100009; j += 2 * i) {
arr[j] = 1;
}
}
}
arr[1] = 1;
for... |
Problem: You've got an n × 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[] primes = new boolean[1000001];
public static void sieve() {
Arrays.fill(primes, true);
primes[0] = false;
primes[1] = false;
for (int i = 2; i < primes.length; i++) {
if (!primes[i]){continue;}
f... |
Problem: You've got an n × 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... | lm=100018;p=[1,1]+[0]*lm
for i in range(2,lm):p[i*i::i]=[1]*(lm/i-i+1)
for i in range(lm,0,-1):p[i]*=p[i+1]+1
I=lambda _:map(int,raw_input().split());n,m=I(0);M=map(I,[0]*n)
print min(sum(p[i]for i in r)for r in M+zip(*M)) |
Problem: You've got an n × 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... | # aadiupadhyay
import os.path
from math import gcd, floor, ceil
from collections import *
import sys
mod = 1000000007
INF = float('inf')
def st(): return list(sys.stdin.readline().strip())
def li(): return list(map(int, sys.stdin.readline().split()))
def mp(): return map(int, sys.stdin.readline().split())
def inp(): re... |
Problem: You've got an n × 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
numD = 2
prime=[1]*num
prime[1]=0
prime[0]=0
for i in range(numD,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 i... |
Problem: You've got an n × 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() {
vector<int> v;
int isprime[100005];
for (int i = 2; i < 100005; i++) isprime[i] = 1;
isprime[1] = 0;
for (int i = 2; i < 100005; i++) {
if (isprime[i]) {
v.push_back(i);
for (int j = i + i; j < 100005; j += i) isprime[j] = 0;
}
}... |
Problem: You've got an n × 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 {
static boolean[] primes = new boolean[100004];
static TreeSet<Integer> tset = new TreeSet<>();
public static void main(String[] args) {
sieve();
Scanner sc = new Scanner(System.in);
int r=sc.nextInt();
int c =sc.nextInt();
... |
Problem: You've got an n × 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
MAX_P = 100100
is_prime = [True] * MAX_P
primes = []
is_prime[0] = is_prime[1] = False
for i in xrange(2, MAX_P):
if is_prime[i]:
primes.append(i)
for j in xrange(i + i, MAX_P, i):
is_prime[j] = False
n, m = map(int, raw_input().split())
a = [map(int, raw_... |
Problem: You've got an n × 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 main() {
int x[501][501], i, j, ans, k, a[100101], b[32000], num[501][501], n, m, l;
long long int ansx[501], ansy[501], minx, miny;
for (i = 2; i <= 100100; i++) a[i] = 1;
k = 0;
for (i = 2; i <= 500; i++) {
if (a[i] == 1) {
for (j = i * i; j <= 100100; j += i) a[j] = 0... |
Problem: You've got an n × 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,input().split())
s=[[*map(int,input().split())] for _ in " "*n]
limit=int(1e5+2)
l=[1,1]+[0]*limit
for i in range(2,limit):
l[i*i::i]=[1]*((limit-i*i)//i+1)
for i in range(limit,-1,-1):
l[i]*=l[i+1]+1
for i in range(n):
for j in range(m):
s[i][j]=l[s[i][j]]
print(min(min(sum(i) for i ... |
Problem: You've got an n × 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.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int x = input.nextInt();
int y = input.nextInt();
int l;
int g = 200000;
int[][] z = new int[x][y];
int[] ... |
Problem: You've got an n × 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 mxn = 1000001;
int prime[mxn];
void sieve() {
for (long long int i = 0; i < mxn; i++) prime[i] = 0;
prime[0] = prime[1] = 1;
for (int i = 2; i * i < mxn; i++) {
if (prime[i] == 0) {
for (long long int j = i * i; j < mxn; j += i) prime[j] = 1;
}... |
Problem: You've got an n × 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... |
# -*- coding: utf-8 -*-
# @Date : 2019-02-08 08:18:25
# @Author : raj lath (oorja.halt@gmail.com)
# @Link : link
# @Version : 1.0.0
from sys import stdin
max_val=int(10e12)
min_val=int(-10e12)
def read_int() : return int(stdin.readline())
def read_ints() : return [int(x) for x in stdin.readline().spli... |
Problem: You've got an n × 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<bool> prime(2e5, true);
void sieve() {
for (int i = 2; i < prime.size(); i++) {
if (prime[i] == true) {
int k = 2 * i;
while (k < prime.size()) {
prime[k] = false;
k += i;
}
}
}
}
int main() {
sieve();
int n, m;
cin... |
Problem: You've got an n × 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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;
public class Main1 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
//inputs
int n = in.nextInt();
int m = in.nextInt();
... |
Problem: You've got an n × 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... | limite = int(10e5)
primos = [True for i in range(limite)]
primos[0] = False
primos[1] = False
for i in range(2,limite):
if primos[i]:
for j in range(i**2, limite, i):
primos[j] = False
distancias = [0 for i in range(limite)]
distancias[0] = 2
distancias[1] = 1
base = int(limite//10 - 1)
while n... |
Problem: You've got an n × 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):
arr = []
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for p in range(2, n):
if prime[p]:
arr.append(p)
return arr
prime = s... |
Problem: You've got an n × 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... | ######### ## ## ## #### ##### ## # ## # ##
# # # # # # # # # # # # # # # # # # #
# # # # ### # # # # # # # # # # # #
# ##### # # # # ### # # # # # # # # #####
# # # # # # # # # # # # # # # # # #
#########... |
Problem: You've got an n × 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> primes;
int sieveCountPrimesInRange(long long n) {
vector<bool> isPrime(n + 1, true);
int cnt = 0;
isPrime[0] = isPrime[1] = 0;
for (long long i = 1; i <= (n / i); i++) {
if (isPrime[i])
for (long long j = i * 2; j <= n; j += i) isPrime[j] = 0;... |
Problem: You've got an n × 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 MOD = 1e6 + 3;
const int INFI = 1e9 * 2;
const int N = 555;
const int M = 111111;
const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};
bool p[M];
int next(int n) {
int m = n;
while (p[m]) m++;
return m - n;
}
int r[N], c[N];
int mai... |
Problem: You've got an n × 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 long double pi = 3.14159265358979323;
const double EPS = 1e-12;
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
vector<long long> v;
void primearray() {
bool prime[N];
memset(prime, 1, sizeof(prime));
prime[0] = 0;
prime[1] = 1;
for (int i = 2; i < sqrt(N); ... |
Problem: You've got an n × 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() {
int pr[110000];
int prev = 0;
pr[0] = pr[1] = 2;
for (int i = 2; i < 110000; i++) {
bool prr = true;
for (int j = 2; j * j <= i; j++)
if (i % j == 0) {
prr = false;
break;
}
if (prr) {
for (int j = prev; j <... |
Problem: You've got an n × 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 = int(10e5)
crivo = [True for i in range(ncrivo)]
crivo[0] = False
crivo[1] = False
for i in range(2, ncrivo):
if crivo[i]:
for j in range(i ** 2, ncrivo, i):
crivo[j] = False
# frequencia
contador = [0 for i in range(ncrivo)]
contador[100000] = 3
for i in range(99999, -1, -1):
if c... |
Problem: You've got an n × 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... | MX = 100500
prime = [True for i in xrange(MX+10)]
ds = [0 for i in xrange (MX+10)]
a = [[0 for i in xrange (510)] for j in xrange (510)]
def sieve():
prime[0] = prime[1] = False
for i in range (3,MX,2):
if (prime[i] == True):
for j in range (3*i,MX,2*i):
prime[j] = False
... |
Problem: You've got an n × 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 arr[1000000 + 100];
void findPrime(int size) {
memset(arr, true, sizeof(arr));
arr[0] = false;
arr[1] = false;
for (int i = 2; i <= size; i++)
if (arr[i])
for (int j = i + i; j <= size; j += i) arr[j] = false;
}
int main() {
int n, m;
while (cin >... |
Problem: You've got an n × 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 B {
boolean primes[] = new boolean[100091];
public void generate(int n){
primes[0] = true;
primes[1] = true;
for(int i = 2 ; i <= (int)Math.sqrt(n) ; i++){
for(int j=i+1 ; j<=n ; j++){
if( !primes[j] && j%i==0 ) primes[j] = true;
}
}
... |
Problem: You've got an n × 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 get_input_matrix():
user_input = input().split()
n = int(user_input[0])
m = int(user_input[1])
matrix = []
for i in range(n):
user_input_matrix = input().split()
row = []
for j in range(m):
row.append(int(user_input_matrix[j]))
matrix.append(row)
... |
Problem: You've got an n × 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 PA {
/*
Tyger! Tyger! burning bright
In the forests of the night,
What immortal hand or eye
Could frame thy fearful symmetry?
In what distant deeps or skies
Burnt the fire of thine eyes?
On what wings dare... |
Problem: You've got an n × 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 × 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... | # Mateus Brito de Sousa Rangel - 117110914
limit = 300000
primes = [False for x in range(limit)]
columns = [0 for x in range(limit)]
ans = limit
Input = map(int, input().split())
l, c = list(Input)
def crivo():
primes[0] = True
primes[1] = True
for i in range(2, limit):
if not primes[i]:
... |
Problem: You've got an n × 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
MAX = 10**5+10
x =[True] * MAX
for i in range(2, MAX):
if x[i]:
for j in range(i*2, MAX, i):
x[j] = False
ps = [i for i in range(2, MAX) if x[i]]
I = lambda:map(int, raw_input().split())
n,m=I()
r,c=[0]*n,[0]*m
for i in range(n):
a = I()
for j in range(m):
k = bisec... |
Problem: You've got an n × 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... | #!/usr/bin/python3
n, m = tuple(map(int, input().split()))
a = [list(map(int, input().split())) for _ in range(n)]
simple = [0] * (10**5 + 4)
simple[1] = 1
for i in range(2, 10**5 + 4):
if simple[i] == 0:
j = 2
while i * j < 10**5 + 4:
simple[i * j] = 1
j += 1
simple[-1] = ... |
Problem: You've got an n × 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 {
static boolean[] p = new boolean[1000001];
public static void sieve() {
Arrays.fill(p, true);
p[0] = false;
p[1] = false;
for (int i = 2; i <= 1000000; i++) {
if (!p[i]) {
continue;
}
... |
Problem: You've got an n × 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.*;
/**
*
* @author Togrul Gasimov (ttogrul30@gmail.com)
* Created on 13.09.2013
*/
public class Main {
public static void main(String[] args) /*throws FileNotFoundException*/ {
InputStream inputStream = System.in;
OutputStream outputStream = Syste... |
Problem: You've got an n × 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.*;
import java.util.stream.*;
public class P271B {
public BitSet genPrimes(int n) {
long [] lPrimes = new long [n / Long.SIZE + 1];
Arrays.fill(lPrimes, 0xAAAAAAAAAAAAAAAAL);
BitSet primes = BitSet.valueOf(lPrimes);
primes.flip(1, 3);
primes... |
Problem: You've got an n × 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... | # Generate array of primes using Sieve of Eratosthenes
def arrayOfPrimes():
n = 100000
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
prime[0] = False
prime[1] = False
return prime
# Generate... |
Problem: You've got an n × 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... | enter = input().split(" ")
rows = int(enter[0])
columns = int(enter[1])
matrix = []
minMove = 100000
def is_prime(n: int):
if n <= 3:
return n > 1
if n % 2 == 0 or n % 3 == 0:
return False
i = 5
while i ** 2 <= n:
if n % i == 0 or n % (i + 2) == 0:
return False
... |
Problem: You've got an n × 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
max_num = 110005
max_range = 505
n, m = map(int, raw_input().split())
d = [0 for i in xrange(max_num)]
is_prime = [True for i in xrange(max_num)]
is_prime[1] = False
for i in xrange(2, 110001):
if is_prime[i]:
for j in xrange(i+i, 110001, i):
is_prime[j] = False
for i in xrange(11... |
Problem: You've got an n × 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
prm=[1 for i in range(101001)]
prm[0]=0
prm[1]=0
for i in range(2,int(math.sqrt(101001))+1):
if prm[i]==1:
for j in range(i*i,101001,i):
prm[j]=0
n,m=map(int,input().split())
arr=[]
row=[0]*n
col=[0]*m
for i in range(n):
l=list(map(int,input().split()))
for k in range(m):
... |
Problem: You've got an n × 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... | isPrime = [1] * 100010
def Prime():
global isPrime
isPrime[0] = isPrime[1] = 0
for i in range(100010):
if isPrime[i]:
for j in range(2 * i, 100010, i):
isPrime[j] = 0
Prime()
n, m = map(int, input().split())
l = []
for i in range(n):
l.append([int(x) for x in input()... |
Problem: You've got an n × 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;
set<int> primes;
int sumr[500];
int sumc[500];
void addpr(int n) {
bool b;
int i, j;
primes.insert(2);
for (i = 3; i <= n; i += 2) {
b = true;
j = 3;
while (b && j * j <= i) {
b = i % j != 0;
j += 2;
}
if (b) primes.insert(i);
}
}
i... |
Problem: You've got an n × 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.StringTokenizer;
public class PrimeMatrix {
public static void main(String[] args) {
boolean[] isComp = new boolean[200002];
int[] v = new int[200002];
i... |
Problem: You've got an n × 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 {
int INF = Integer.MAX_VALUE / 1000;
static Scanner sc = null;
int MAX = 200001;
public void solve() throws Exception{
int n = sc.nextInt();
int m = sc.nextInt();
int[][] d = new int[n][m];
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){... |
Problem: You've got an n × 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 n, m;
int matrix[500][500];
int min = -1;
short primeMap[100500];
bool calcIsPrime(int n) {
if (n < 2) return false;
for (int i = 2; i <= sqrt(n * 1.); i++) {
if (n % i == 0) return false;
}
return true;
}
void genPrimes() {
primeMap[1] = -1;
for (int i ... |
Problem: You've got an n × 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())
g = [[] for i in xrange(n)]
for i in xrange(n):
g[i] = map(int, raw_input().split())
p = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,2... |
Problem: You've got an n × 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, n = map(int, input().split(' '))
X = list(zip(*[map(int,input().split()) for i in [True] * m]))
from bisect import bisect_left as bsl
MaxPrime = 100004
def vec_primes(n): # See exercise 35.
""" Returns a list of primes < n """
sieve = [True] * (n//2)
for i in range(3,int(n**0.5)+1,2):
if siev... |
Problem: You've got an n × 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<long long> v(1e5 + 1, -1);
bool isprime(int t) {
if (t == 2) return true;
if (!(t % 2)) return false;
for (int x = 3; x * x <= t; x += 2) {
if (!(t % x)) return false;
}
return true;
}
int main() {
int a, mn = INT_MAX, s;
cin >> a >> s;
v[1] = 1;
... |
Problem: You've got an n × 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... | #atal 2020.1, feito baseado nas noocoes vistas, conceito do crivuuuuus facilment eencontrado.
import sys
import math
limit = 100025
minhaListadePrimos = [True for i in range(limit + 1)]
primosSeguintestsss = [0 for i in range(200000)]
def crivandu():
minhaListadePrimos[0] = minhaListadePrimos[1] = False
fo... |
Problem: You've got an n × 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, gettrace
import bisect,sys
if not gettrace():
def input():
return next(stdin)[:-1]
# def input():
# return stdin.buffer.readline()
def IP(): # to take tuple as input
return map(int,stdin.readline().split())
def L(): # to take list as input
return list(map(int,stdin.rea... |
Problem: You've got an n × 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 gcd(int a, int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
vector<long long int> adj[100001];
long long int vist[10001];
void dfs(long long int node) {
vist[node] = 1;
for (long long int child : adj[node])
if (!vist[child]) dfs(child);
}
vector<long l... |
Problem: You've got an n × 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 = 600;
long long n, m, a, b, ans, arr[2][N];
vector<long long> prms;
void build() {
const int LIM = 1e6;
bool num[LIM] = {};
for (long long i = 2; i < LIM; ++i) {
if (num[i]) continue;
prms.push_back(i);
for (long long j = i * i; j < LIM; j += ... |
Problem: You've got an n × 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 cmpStruct {
bool operator()(int const& lhs, int const& rhs) const { return lhs > rhs; }
};
long long int power(long long int x, long long int y) {
long long int res = 1;
while (y) {
if (y & 1) res = (res * x) % 1000000009;
y = y / 2, x = (x * x) % 10000... |
Problem: You've got an n × 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 inf = 0x7FFFFFFF;
const double eps = 1e-9L;
const double pi = acos(-1.0);
using namespace std;
int prime[100100 + 10], pf[100100 + 10];
int row[520], col[520], con[100100 + 5];
void make() {
int count = 0;
prime[count++] = 2;
for (int i = 3; i < 100100; i +=... |
Problem: You've got an n × 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) {
Scanner in = new Scanner(System.in);
boolean prime[] = new boolean[100004];
for (int i = 2; i < 50003; i++) {
for (int j = 2; i * j < 100004; j++) {
prime[(i * j) - 1] =... |
Problem: You've got an n × 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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOExcept... |
Problem: You've got an n × 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 generate_primes(n):
primes, sieve = {}, [True] * int(n+1)
for p in range(2, n + 1):
if sieve[p]:
primes[p] = p
for i in range(p * p, n + 1, p):
sieve[i] = False
return primes
def distance_prime(n, primes):
if primes.get(n):
return 0
... |
Problem: You've got an n × 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... | MOD = 1000000007
ii = lambda: int(input())
si = lambda: input()
dgl = lambda: list(map(int, input()))
f = lambda: map(int, input().split())
il = lambda: list(map(int, input().split()))
ls = lambda: list(input())
from bisect import *
l = [1]*(10**5+100)
for i in range(2,int((10**5+100)**0.5)+1):
for j in range(i*i, ... |
Problem: You've got an n × 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;
using namespace std::chrono;
static const int N = 200000;
void getSOE(vector<int>& p) {
vector<bool> v(N + 1, true);
v[0] = v[1] = false;
for (int i = 2; i * i <= N; ++i) {
if (!v[i]) continue;
for (int j = i * i; j <= N; j += i) {
v[j] = false;
}
... |
Problem: You've got an n × 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.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
public class B {
static BufferedReader in;
static StringTokeniz... |
Problem: You've got an n × 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,input().split())
limit=int(1e5+2)
l=[1,1]+[0]*limit
for i in range(2,limit):
l[i*i::i]=[1]*((limit-i*i)//i+1)
for i in range(limit,-1,-1):
l[i]*=l[i+1]+1
s=[[l[j] for j in map(int,input().split())] for _ in ' '*n]
print(min(min(sum(i) for i in s),min(sum(i) for i in zip(*s)))) |
Problem: You've got an n × 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... | row, col = [int(x) for x in raw_input().split()]
matrix = [[int(x) for x in raw_input().split()] for _ in range(row)]
MaxN = 100000+10
u=[1 for i in range(MaxN)]
u[0] = u[1] = 0
for i in range(2,MaxN):
if not u[i]:
continue
j = i
while True:
j += i
if j >= MaxN:
break
... |
Problem: You've got an n × 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<long long int> a(900005, 1);
vector<long long int> v;
void sieve(long long int nn) {
for (long long int i = 2; i < nn; i++) {
if (a[i]) {
v.push_back(i);
for (long long int j = 2; j * i < nn; j++) a[i * j] = 0;
}
}
}
signed main() {
ios_base... |
Problem: You've got an n × 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 char nl = '\n';
const char gg = ' ';
const int M = 1e5 + 5;
bool mark[M];
vector<int> prime;
void sieve() {
int i, j, n;
for (i = 3; i * i <= M; i += 2) {
if (!mark[i]) {
for (j = i * i; j < M; j += i + i) mark[j] = true;
}
}
prime.push_back(2);
... |
Problem: You've got an n × 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 ... |
Problem: You've got an n × 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 crivo(limit):
primes = [True] * limit
primes[0] = False
primes[1] = False
for i in xrange(2, limit):
if (primes[i]):
for j in xrange(i * 2, limit, i):
primes[j] = False
return primes
primes = crivo(10**5 + 100)
n, m = map(int, raw_input().split())
matrix = ... |
Problem: You've got an n × 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... | #!/usr/bin/env python
primes = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,
103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,
199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,
313,317,331,337,347,349,353,359,367,373,379,383,389,397,4... |
Problem: You've got an n × 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 List<Integer> primes = new ArrayList();
static int[] p = new int[110001];
public static void sieve()
{
for(int i=2;i<=110000;i++)
{
if(p[i] == 0) {
for(int j=2;i*j<=110000;j++)
{
p[i*j] = 1;
}
}
}
for(int i=2;i<=110000;i++)if(p[... |
Problem: You've got an n × 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 Sample {
static int MAX = (int)(1e6+2);
static int MOD=(int)1e9+7;
static int countt = 0;
public static void main(String[] args) throws Exception{
// TODO Auto-generated method stub
//BufferedReader br = n... |
Problem: You've got an n × 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.BigInteger;
import java.util.*;
/**
*
* @author Saju
*
*/
public class Main {
private static int dx[] = { -1, 0, 1, 0 };
private static int dy[] = { 0, -1, 0, 1 };
private static final long INF = (long) (1e15);
private static final double EPSILON = 1e-10;
private static ... |
Problem: You've got an n × 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... | a,b=map(int,input().split())
c=100030
f=[0]*c
f[1]=1
for i in range(2,c):
if f[i]==0:
for j in range(i+i,c,i):
f[j]=1
t=0
for i in range(c-1,0,-1):
if f[i]==0:
t=i
f[i]=t
l=[[int(j) for j in input().split()] for i in range(a)]
for i in range(a):
for j ... |
Problem: You've got an n × 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() {
vector<int> prime(100009, 1);
prime[0] = 0;
prime[1] = 0;
for (int i = 2; i * i <= 100009; ++i) {
if (prime[i]) {
for (int j = i * 2; j <= 100009; j += i) {
prime[j] = 0;
}
}
}
int l = 0;
for (int i = 100009; i >= 1; --... |
Problem: You've got an n × 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 L {
static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer st=new StringTokenizer("");
static public String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new... |
Problem: You've got an n × 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;
int t1;
bool M[1000005];
int A[100005];
int rowsum[600], colsum[600];
int main() {
P.push_back(2);
for (int i = 3; i <= 1000; i += 2)
for (int j = i * i; j <= 1e6; j += i) M[j] = 1;
for (int i = 3; i <= 1e6; i += 2) {
if (!M[i]) P.push_back(i);
... |
Problem: You've got an n × 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 fact(long int w) {
long int k, p = 0, l = 0, r = w;
if (w == 1)
return 1;
else {
while (p != 1) {
for (k = 2; k <= pow(w, .5); k++) {
if (w % k == 0) {
l = 1;
k = pow(w, .5);
}
}
if (l == 0)
p ... |
Problem: You've got an n × 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 java.lang.Math.*;
public class Matrix1
{
public static void main(String args[])throws Exception
{
boolean seive[]=new boolean[1000000];
Arrays.fill(seive,true);
BufferedReader br=new BufferedReader(new InputStreamReader(System.i... |
Problem: You've got an n × 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() {
int a[100005];
set<int> b;
for (int i = 2; i <= 100100; i++) {
int flag = 0;
for (int j = 2; j <= sqrt(i); j++) {
if (i % j == 0) flag = 1;
}
if (flag == 0) b.insert(i);
}
set<int>::iterator it, it2;
int r[505], c[505];
memse... |
Problem: You've got an n × 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[100100], i, j, k;
bool b[100100];
void seive() {
a[0] = 2;
a[1] = 2;
a[2] = 2;
for (i = 4; i <= 100100; i += 2) b[i] = 1;
for (i = 3; i * i <= 100100; i += 2) {
if (b[i] == 0)
for (j = i * i; j <= 100100; j += i) b[j] = 1;
}
for (j = 100100; j ... |
Problem: You've got an n × 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 pointonline{
static int n,k;
static StringBuilder ans;
static HashMap<Integer,Integer> map=new HashMap<>();
public static void main(String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
... |
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