Zero Wind - Jamie Wong » pascal http://jamie-wong.com Inside the mind of a Waterloo Software Engineering student Sun, 15 Aug 2010 05:25:49 +0000 en hourly 1 http://wordpress.org/?v=3.0 SPOJ Problem 2: Prime Generator (PRIME1) http://jamie-wong.com/2009/11/12/spoj-problem-2-prime-generator/ http://jamie-wong.com/2009/11/12/spoj-problem-2-prime-generator/#comments Fri, 13 Nov 2009 05:47:34 +0000 Jamie Wong http://jamie-wong.com/me/?p=47 Problem: Prime Generator

Peter wants to generate some prime numbers for his cryptosystem. Help him! Your task is to generate all prime numbers between two given numbers!

Concept The idea behind every solution here (with some variation) is to generate all the prime numbers that could be factors of numbers up to the maximum endpoint 1 billion. That square root happens to be around 32000. Using this array, do a bounded Sieve of Eratosthenes only in the range requested. In languages like php and python, it turns out that it's more efficient to build an associative array and check if the index is set than it is to generate a huge boolean array.

Code

C++

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68

/// PRIME1 - C++ (g++)
// AC Time: 2.52s
// NOTE: I am aware that the use of vector and set actually
//  makes this code run _slower_
// I used vector and set simply as a way of practicing STL
#include <iostream>
#include <cmath>
#include <vector>
#include <set>
using namespace std;
 
int main() {
    vector<int> primes;
    primes.push_back(2);
 
    for (int i = 3; i <= 32000; i+=2) {
        bool isprime = true;
        int cap = sqrt(i) + 1;
 
        vector<int>::iterator p;
        for (p = primes.begin(); p != primes.end(); p++) {
            if (*p >= cap) break;
            if (i % *p == 0) {
                isprime = false;
                break;
            }
        }
        if (isprime) primes.push_back(i);
    }
 
    int T,N,M;
 
    cin >> T;
 
    for (int t = 0; t < T; t++) {
        if (t) cout << endl;
 
        cin >> M >> N;
        if (M < 2) M = 2;
 
        int cap = sqrt(N) + 1;
 
        set<int> notprime;
        notprime.clear();
 
        vector<int>::iterator p;
        for (p = primes.begin(); p != primes.end(); p++) {
 
            if (*p >= cap) break;
            int start;
 
            if (*p >= M) start = (*p)*2;
            else start = M + ((*p - M % *p) % *p);
 
            for (int j = start; j <= N; j += *p) {
                notprime.insert(j);
            }
        }
 
        for (int i = M; i <= N; i++) {
            if (notprime.count(i) == 0) {
                cout << i << endl;
            }
        }
 
    }
    return 0;
}

C99

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64

// PRIME1 - C99 (gcc)
// AC Time: 0.07s
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>
 
int main() {
    int primes[4000];
    int numprimes = 0;
 
    primes[numprimes++] = 2;
    for (int i = 3; i <= 32000; i+=2) {
        bool isprime = true;
        int cap = sqrt(i)+1;
        for (int j = 0; j < numprimes; j++) {
            if (primes[j] >= cap) break;
            if (i % primes[j] == 0) {
                isprime = false;
                break;
            }
        }
        if (isprime) primes[numprimes++] = i;
    }
 
    int T,N,M;
    scanf("%d",&T);
 
    for (int t = 0; t < T; t++) {
        if (t) printf("\n");
        scanf("%d %d",&M,&N);
        if (M < 2) M = 2;
 
        int cap = sqrt(N) + 1;
 
        bool isprime[100001];
        memset(isprime,true,sizeof(isprime));    
 
        for (int i = 0; i < numprimes; i++) {
            int p = primes[i];
 
            if(p >= cap) break;
 
            int start;
 
            if (p >= M) start = p*2;
            else start = M + ((p - M % p) % p);
 
            for (int j = start; j <= N; j += p) {
                isprime[j - M] = false;
            }
        }
 
        int start = (M % 2)?M:M+1;    
 
        if (M == 2) {
            printf("2\n");
        }
        for (int i = start; i <= N; i+=2) {
            if (isprime[i-M]) printf("%d\n",i);
        }
    }
    return 0;
}

PHP

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63

<?
// PRIME1 - PHP
// AC Time: 5.73
 
$primes = array(2);
$numprimes = 1;
 
for ($i = 3; $i <= 32000; $i+=2) {
    $isprime = true;
    $cap = sqrt($i)+1;
 
    for ($j = 0; $j < $numprimes; ++$j) {
        if ($j >= $cap) break;
        if (!($i % $primes[$j])) {
            $isprime = false;
            break;
        }
    }
    if ($isprime) $primes[$numprimes++] = $i;
}
 
fscanf(STDIN,"%d",$T);
 
$output = "";
 
for ($t = 0; $t < $T; $t++) {
    if ($t) $output .= "\n";
 
    fscanf(STDIN,"%d %d",$M,$N);
    if ($M < 2) $M = 2;
 
    $isprime = array();
 
    $cap = sqrt($N)+1;
 
    $i = 0;
    while($i < $numprimes) {
        $p = $primes[$i];
 
        if ($p >= $cap) break;
 
        if ($p >= $M) {
            $start = $p*2;
        } else {
            $start = $M + (($p - $M % $p) % $p);
        }
 
        for ($j = $start; $j <= $N; $j += $p) {
            $isprime[$j - $M] = false;
        }
 
        ++$i;
    }
 
    for ($i = $M; $i <= $N; ++$i) {
        if (!isset($isprime[$i-$M])) {
            $output .= "$i\n";
        }
    }
}
 
echo $output;
?>

Python

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59

# PRIME1 - Python
# AC Time: 3.10s
from math import sqrt
primes = [2]
 
for i in range(3,32000,2):
    isprime = True
 
    cap = sqrt(i)+1
 
    for j in primes:
        if (j >= cap):
            break
        if (i % j == 0):
            isprime = False
            break
    if (isprime):
        primes.append(i)
 
T = input()
output = ""
for t in range(T):
 
    if (t > 0):
        output += "\n"
 
    M,N = raw_input().split(' ')
    M = int(M)
    N = int(N)
    cap = sqrt(N)+1
 
    if (M < 2):
        M = 2
 
    isprime = [True]*100001
 
    for i in primes:
        if (i >= cap):
            break
 
        if (i >= M):
            start = i*2
        else:
            start = M + ((i - M % i)%i)
 
        # The two below, obscure lines create a continuous
        #  block of false elements in order to set all
        #  elements correspnding to numbers divisible by i
        #  in isprime to be false
        # In turns out that this runs substantially faster
        #  than setting the elements individually using loops
        falseblock = [False] * len(isprime[start-M:N+1-M:i]);
        isprime[start-M:N+1-M:i] = falseblock
 
    for i in range(M,N+1):
        if (isprime[i-M] == True):
            output += str(i) + "\n"
 
print output[:-1]

Bash

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35

# PRIME1 Incomplete - BASH
# This is not a complete solution
# The speed of bash parsing makes getting an 
#  AC submission infeasible
 
# The following code is a working prime generator
# Giving enough time, it will output all prime 
#  numbers from 0 to 32000
# ..which is the first step in the solution to PRIME1
 
let PRIMES[0]=2
let NUMPRIMES=1
for i in {3..32000}; do
    let ISPRIME=1
 
    #for j in {0..$NUMPRIMES}; do
    for (( j = 0;j<$NUMPRIMES;j++ )); do
        let CURPRIME=${PRIMES[$j]}
        if [[ $CURPRIME*$CURPRIME -gt $i ]]; then
            break
        fi
 
        let MOD=$[$i % $CURPRIME]
        if [ $MOD -eq 0 ]; then
            let ISPRIME=0
            break
        fi
    done
 
    if [ $ISPRIME -eq 1 ]; then
        echo $i
        let PRIMES[NUMPRIMES]=$i
        let NUMPRIMES=NUMPRIMES+1
    fi
done

Ruby

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80

# PRIME1 - Ruby
# AC Time: 4.65s
# 
# While I would have liked to use more
# Ruby idioms such as for p in primes
# or primes.each do |p| (used once),
# the difference in runtime between
# the use of loops like these and while
# loops was non-negligible
 
primes = [2]
 
3.step(32000,2) do |i|
    isprime = true
    cap = Math.sqrt(i) + 1
 
    primes.each do |p|
        if (p >= cap)
            break
        end
 
        if (i % p == 0)
            isprime = false
            break
        end
    end
 
    if isprime
        primes << i
    end
end
numprimes = primes.length
 
T = gets.to_i
 
 
output = ""
t = 0
while t < T
    print "\n" if t > 0
    line = gets.split(" ")
    m = line[0].to_i
    n = line[1].to_i
 
    m = 2 if m < 2
 
    cap = Math.sqrt(n) + 1
 
    notprime = {}
 
    i = 0
    while i < numprimes
        p = primes[i]
        i+=1
        if (p >= cap)
            break
        end
 
        if (p >= m)
            start = p*2
        else 
            start = m + ((p - m % p)%p)
        end
 
        j = start
        while j <= n
            notprime[j] = true
            j += p
        end
    end
 
    i = m
    while (i <= n)
        if (notprime[i] == nil)
            print i,"\n"
        end
        i+=1
    end
    t += 1
end

Java

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65

// PRIME1 - Java
// AC Time: 2.20
 
import java.io.*;
import java.util.*;
import java.lang.Math.*;
 
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
 
        int[] primes = new int[4000];
        int numprimes = 0;
 
        primes[numprimes++] = 2;
        for (int i = 3; i <= 32000; i+=2) {
            boolean isprime = true;
            double cap = Math.sqrt(i) + 1.0;
 
            for (int j = 0; j < numprimes; j++) {
                if (j >= cap) break;
                if (i % primes[j] == 0) {
                    isprime = false;
                    break;
                }
            }
            if (isprime) primes[numprimes++] = i;
        }
 
 
        int T,N,M;
 
        T = in.nextInt();
 
        for (int t = 0; t < T; t++) {
            if (t > 0) System.out.println("");
 
            M = in.nextInt();
            N = in.nextInt();
 
            if (M < 2) M = 2;
 
            boolean[] isprime = new boolean[100001];
            for (int j = 0; j < 100001; j++) {
                isprime[j] = true;
            }
 
            for (int i = 0; i < numprimes; i++) {
                int p = primes[i];
                int start;
 
                if (p >= M) start = p*2;
                else start = M + ((p - M % p)%p);
 
                for (int j = start; j <= N; j += p) {
                    isprime[j - M] = false;
                }
            }
 
            for (int i = M; i <= N; i++) {
                if (isprime[i-M]) System.out.println(i);
            }
        }
    }
}

Perl

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64

# PRIME1 - Perl
# AC Time: 4.28s
@primes = (2);
 
for ($i = 3; $i <= 32000; $i+=2) {
    $isprime = 1;
    $cap = sqrt($i) + 1;
    foreach $p (@primes) {
        if ($p >= $cap) { 
            last;
        }
        if ($i % $p == 0) {
            $isprime = 0;
            last;
        }
    }
    if ($isprime == 1) {
        push(@primes,$i)
    }
}
 
 
$T = <STDIN>;
 
for ($t = 0; $t < $T; $t++) {
    if ($t > 0) {
        print "";
    }
 
    $in = <STDIN>;
    @line = split(/ /,$in);
    $M = $line[0];
    $N = $line[1];
 
    if ($M < 2) {
        $M = 2;
    }
 
    $cap = sqrt($N) + 1;
 
    @isprime = ((1) x 100001);
 
    foreach $p (@primes) {
        if ($p >= $cap) {
            last;
        }
 
        if ($p >= $M) {
            $start = $p*2;
        } else {
            $start = $M + (($p - $M % $p) % $p);
        }
 
        for ($j = $start; $j <= $N; $j += $p) {
            $isprime[$j-$M] = 0;
        }
    }
 
    for ($i = $M; $i <= $N; $i++) {
        if ($isprime[$i-$M] == 1) {
            print "$i\n";
        }
    }
}

C#

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61

// PRIME1 - C# (gmcs)
// AC Time: 1.50s
 
using System;
class PRIME1 {
    static void Main() {
 
        int[] primes = new int[4000];
        int numprimes = 0;
 
        primes[numprimes++] = 2;
        for (int i = 3; i <= 32000; i+=2) {
            bool isprime = true;
            double cap = Math.Sqrt(i) + 1.0;
            for (int j = 0; j < numprimes; j++) {
                if (j >= cap) break;
                if (i % primes[j] == 0) {
                    isprime = false;
                    break;
                }
            }
            if (isprime) primes[numprimes++] = i;
        }
 
        int T,N,M;
        T = int.Parse(Console.ReadLine());
 
        for (int t = 0; t < T; t++) {
            if (t > 0) Console.WriteLine("");
 
            string[] parts = Console.ReadLine().Split(' ');
            M = int.Parse(parts[0]);
            N = int.Parse(parts[1]);
 
 
            if (M < 2) M = 2;
            double cap = Math.Sqrt(N)+1;
 
            bool[] isprime = new bool[100001];
            for (int i = 0; i < 100001; i++) isprime[i] = true;
 
            for (int i = 0; i < numprimes; i++) {
                int p = primes[i];
                if (p >= cap) break;
                int start;
 
 
                if (p >= M) start = p*2;
                else start = M + ((p - M % p)%p);
 
                for (int j = start; j <= N; j+= p) {
                    isprime[j - M] = false;
                }
            }
 
            for (int i = M; i <= N; i++) {
                if (isprime[i-M] == true) Console.WriteLine(i);
            }
        }
    }
}

GNU Pascal

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90

{ 
    PRIME1 - GNU Pascal
    AC Time: 0.54
}
 
program PRIME1;
var
    primes: array[1..4000] of integer;
    numprimes: integer;
    i,j: integer;
    cap: double;
    isprime: boolean;
    T,N,M: integer;
    isp: array[0..100001] of boolean;
    p,start: integer;
 
 
begin
    primes[1] := 2;
    numprimes := 1;
    for i := 3 to 32000 do
    begin
        isprime := true;
        cap := sqrt(i) + 1;
        for j := 1 to numprimes do
        begin
            if primes[j] >= cap then 
                    break;
 
            if (i MOD primes[j] = 0) then
            begin
                isprime := false;
                break;
            end
        end;
 
        if isprime = true then 
        begin
            numprimes := numprimes + 1;
            primes[numprimes] := i
        end
    end;
 
    read(T);
 
    for t := 1 to T do
    begin
        if t > 0 then
            writeln('');
 
        read(M);
        read(N);
 
        if M < 2 then
            M := 2;
 
        cap := sqrt(N) + 1;
 
        for i := 0 to 100001 do
            isp[i] := true;
 
 
        for i := 1 to numprimes do
        begin
            p := primes[i];            
            if p >= cap then 
                break;
 
            if p >= M then
                start := (p * 2);
 
            if p < M then
                start := M + ((p - (M MOD p)) MOD p);
 
            j := start;
 
            while j <= N do
            begin
                isp[j - M] := false;
                j := j + p;
            end;
        end;
 
        for i := M to N do
        begin
            if isp[i-M] = true then
                writeln(i);
        end;
    end;
end.
]]> http://jamie-wong.com/2009/11/12/spoj-problem-2-prime-generator/feed/ 4 SPOJ Problem 1 http://jamie-wong.com/2009/11/08/spoj-problem-1/ http://jamie-wong.com/2009/11/08/spoj-problem-1/#comments Sun, 08 Nov 2009 17:55:10 +0000 Jamie Wong http://jamie-wong.com/me/?p=41 When I saw that Thor had done solutions to the first few problems of Project Euler in many many languages, I thought "Hey! That's a good idea!" but didn't want to copy exactly.

So instead I'm doing solutions to some problems on SPOJ. SPOJ is an online judge system full of algorithmic problems. This is one of the things that Hanson Wang did to get as good as he is. The solutions I'll be providing here are going to be very simple problems, so don't expect any magic. The beautiful thing about SPOJ is the sheer number of languages it will judge. I figured this was the perfect playground to make sure my code worked in all the languages I try.

Problem: Life, the Universe, and Everything

Your program is to use the brute-force approach in order to find the Answer to Life, the Universe, and Everything. More precisely... rewrite small numbers from input to output. Stop processing input after reading in the number 42. All numbers at input are integers of one or two digits.

Solutions - In order of frequency that I use the language

C++

1
2
3
4
5
6
7
8
9
10
11
12
13

//TEST AC - CPP (g++)
#include <iostream>
using namespace std;
 
int main() {
	int n;
	while(1) {
		cin >> n;
		if (n == 42) break;
		cout << n << endl;
	}
	return 0;
}

C99

1
2
3
4
5
6
7
8
9
10
11

//TEST AC - (gcc C99)
#include <stdio.h>
int main() {
	int n;
	while(1) {
		scanf("%d",&n);
		if (n == 42) break;
		printf("%d\n",n);
	}
	return 0;
}

php

1
2
3
4
5
6
7
8

<?
//TEST AC - PHP
while(1) {
	fscanf(STDIN,"%d",$n);
	if ($n == 42) break;
	print "$n\n";
}
?>

Python

1
2
3
4
5
6

#TEST AC - Python
while 1:
	num = input()
	if (num == 42):
		break
	print num

Bash

1
2
3
4
5
6
7
8
9
10

#!/bin/bash
# TEST AC - BASH
 
while true; do
	read n
	if [ $n -eq 42 ]; then
		break
	fi
	echo "$n"
done

Ruby

1
2
3
4
5
6
7
8

#TEST AC - Ruby
while 1
	n = gets.to_i
	if n == 42
		break
	end
	puts n
end

Java

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15

//TEST AC - Java
import java.io.*;
import java.util.*;
 
public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
 
		while(true) {
			int n = in.nextInt();
			if (n == 42) break;
			System.out.println(n);
		}
	}
}

Perl

1
2
3
4
5
6
7
8
9

#TEST AC - Perl
while (1)
{
	$n = <STDIN>;
	if ($n == 42) {
		last;
	}
	print $n
}

C#

1
2
3
4
5
6
7
8
9
10
11
12

//TEST AC - C# (gmcs + Mono)
using System;
class WelcomeCSS {
	static void Main() {
		while(true) {
			int n;
			n = int.Parse(Console.ReadLine());
			if (n == 42) break;
			Console.WriteLine(n);
		}
	}
}

GNU Pascal

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

{TEST AC - GPC Pascal}
 
program TEST;
 
var n:integer;
 
begin
	while true do
		begin
			readln(n);
			if n = 42 then begin
				break;
			end;
			writeln(n);
		end;
end.

You can see Thor's Project Euler solutions here: Derek Thurn's Website - Project Euler

]]> http://jamie-wong.com/2009/11/08/spoj-problem-1/feed/ 0