算法，在给定的最小值和最大值之间找到“数字不除以平方数”

Question

Problem

There is a minimum value min and maximum value max, find all the numbers which not divided by "square numbers" between min and max. (1<=min<=1trilion, max-min<=1,000,000)

Example

min is 1, max is 10.

The answer is 7 from 1,2,3,5,6,7,10 because 4,8 is divided by square number 2^2=4.

---

My approach.

1. All numbers divided by square number are also divided by square prime number.

2. Using seive of eratosthenes, computed all the square prime numbers smaller than sqrt(max)

3. Finding all numbers divided by above suqare prime numbers.

But I got "time limit" or "Wrong" from that site. How can I prove? Below is code.

void Eratos()
{
    for (ll i = 2; i <= primeMax; i++) num[i] = i;

    for (ll i = 2; i <= 100; i++) {
        for (ll j = 2; j <= primeMax; j++) {
            if (num[j] == -1) continue;
            else if (num[j] > i && num[j] % i == 0) num[j] = -1;
        }
    }

    for (ll i = 2, j = 0; i <= primeMax; i++) if (num[i] != -1) {
        prime[j] = num[i]*num[i];
        j++;
    }
}

Answer 1

You are checking check all numbers in range (nr=max-min) against all primes (np), so complexity is O(nr*np) .

But instead you can make list of prime squares and use Eratosthenes sieve approach again over min..max range, marking squareful numbers. So code will make about

Sum (nr / (sq[0]) + nr / (sq[1]) +...nr / (sq[np-1])

steps with complexity (perhaps) about O(np + nr)

For example, range is 50...80. Prime square list is 4,9,25,49. Make boolean or bit array with 31 entries. First run marks entries 52,56,60..80. Second run: 54, 63, 72; third run marks 50,75, and fourth marks nothing. Now unmarked entries 51,53,55..79 is what you need.

The first entry in range, divisible by some square psq is

((min + psq - 1) div psq) * psq

Delphi code gives result 607923 in 0.1 second

function CountSquareless(AMin, AMax: Int64): Integer;

var
  PrSqList: TList<Int64>;

procedure MakePrimeList; //Eratosphenes sieve
var
  num: array of Byte;
  i, j: Integer;
begin
  SetLength(num, 1000001);
  PrSqList := TList<Int64>.Create;
  for i := 2 to 1000000 div 2 do
    for j := 2 to (1000000 div i) do
       num[i * j] := 1;
  for i := 2 to 1000000 do
    if num[i] = 0 then
       PrSqList.Add(Int64(i) * i);
end;

var
  num: array of Byte;
  i, nr: Integer;
  psq, first: Int64;
begin
  nr := AMax - AMin;
  SetLength(num, nr + 1);
  MakePrimeList;

  for psq in PrSqList do begin
    first := ((AMin + psq - 1) div psq) * psq;
    while first < AMax do begin //Eratosphenes-like sieve uses prime squares
      num[first - AMin] := 1;
      first := first + psq;
    end;
  end;

  Result := 0;
  for i := 0 to nr - 1 do
    if num[i] = 0 then
      Result := Result + 1;
end;

procedure TForm1.Button1Click(Sender: TObject);
begin
  Memo1.Lines.Add(IntToStr(CountSquareless(1000000000000 - 1000000, 1000000000000)));

c++ code for sieve

using namespace std;

typedef long long ll;
const int primeMax = 1000000;
int num[primeMax + 1];
ll prime[100000];
int primecnt = 0;

void Eratos()
{
    for (int i = 2; i <= primeMax; i++) num[i] = i;

    for (int i = 2; i <= primeMax; i++) {
        for (int j = 2; j <= primeMax / i ; j++) {
             num[i * j] = -1;
        }
    }

    for (int i = 2; i <= primeMax; i++)
        if (num[i] != -1) {
          prime[primecnt++] = (ll)num[i] * num[i];
      }
    cout << "last prime squared " << prime[primecnt-1] << " number " << primecnt;
 }

Answer 2

Given that the problem statement asks you to "find all the numbers which not divided" (sic), the fact that you only have one printf and it's not inside a loop of some sort, seems to indicate your solution is more on the "wrong" side of the ledger rather than the "time limit" side.

I'd suggest fixing that before you worry about speeding it up. You can't get any less optimised than "wrong" :-)

---

The first thing you should have done is to see what output you got from the input given in the test case - you would have immediately seen that speed is not the issue here:

pax> echo 1 10 | ./testprog
7

Now my eyes might be getting a bit old nowadays but even I can see there's a difference between 7 and 1,2,3,5,6,7,10 .