Prime triplets (Project Euler 196)

original version

hackerrank programming version

题目大意是定义了一个正整数的表，第一行是1，第二行是1,2，第三行1,2,3...定义prime triple是在表上八连通的三个质数。然后问某行有多少个质数至少在一个prime triple中。   行数 <= 1e7.

题解：

假设要求第n行的答案，只要把上2行和下2行的数都抠出来，然后判断一下就好了。问题转化为求[L,R]之间的质数，这里5行数，区间长度大概有5e7，所以要用modified 区间筛法。 参考 http://euler.stephan-brumme.com/196/ 。 没什么思维量，积累一个区间筛法模板。

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int mod = 1e9 + 7;

char* isPrime;
LL bias;

vector<LL> eratosthenesOddSingleBlock(LL from, LL to)
{
    vector<LL> primes; 
    if (from > to || to <= 1) return primes;
    if (to == 2) 
    {
        primes.push_back(2);
        return primes;
    }
    if (!(from & 1)) ++from;
    if (from > to) return primes;
 
    if (from <= 2) primes.push_back(2), from = 3;
    const int memorySize = (to - from) / 2 + 1;
    char* isPrime = new char[memorySize];

  
    for (int i = 0; i < memorySize; i++)
        isPrime[i] = 1;
    for (LL i = 3; i*i <= to; i+=2)
    {
      if (i >= 3*3 && i % 3 == 0)
         continue;
      if (i >= 5*5 && i % 5 == 0)
         continue;
      if (i >= 7*7 && i % 7 == 0)
         continue;
      if (i >= 11*11 && i % 11 == 0)
         continue;
      if (i >= 13*13 && i % 13 == 0)
         continue;
      LL minJ = ((from+i-1)/i)*i;
      if (minJ < i*i)
          minJ = i*i;
      if ((minJ & 1) == 0)
          minJ += i;
      for (LL j = minJ; j <= to; j += 2*i)
      {
          int index = j - from;
          isPrime[index/2] = 0;
      }
    }

    for (int i = 0; i < memorySize; i++)
           if (isPrime[i]) primes.push_back(from + 2*i);

    delete[] isPrime;
    return primes;
}


bool checkPrime(LL val)
{
    return isPrime[val - bias];
}

inline LL getNumber(int r, int c)
{
    return 1LL * (r - 1) * r / 2 + c;
}

bool checkCentre(int r, int c)
{
    if (c < 0 || c > getNumber(r, c) || !checkPrime(getNumber(r, c))) return false;
    int cnt = 0, x, y;
    for (int dx = -1; dx <= 1; ++dx)
    {
        for (int dy = -1; dy <= 1; ++dy)
        {
            x = r + dx;
            y = c + dy;
            if (y < 0 || y > x) continue;
            cnt += checkPrime(getNumber(x, y));
            if (cnt >= 3) return true;
        }
    }
    return false;
}
bool check(int r, int c)
{
    for (int dx = -1; dx <= 1; ++dx)
        for (int dy = -1; dy <= 1; ++dy)
            if (checkCentre(r + dx, c + dy))
               return true;
    return false;
}

LL solve(int row)
{
    if (row == 1) return 0;
    if (row <= 3) return 5;
    
    LL l = getNumber(row - 2, 1), r = getNumber(row + 2, row + 2);
    vector<LL> primes = eratosthenesOddSingleBlock(l, r);
    isPrime = new char[r - l + 1];
    memset(isPrime, 0, r - l + 1);
    bias = l;
    for (auto p: primes) isPrime[p - bias] = 1;
    
    pair<int, int> pos;
    LL res = 0, val = getNumber(row, 1);
    for (int col = 1; col <= row; ++col)
    {
        if (checkPrime(val) && check(row, col)) 
           res += val;
        ++val;
    }
    return res;
}

int main()
{
    //freopen("out.txt", "w" , stdout);
    
    int a, b;
    cin >> a >> b; 
    cout << solve(a) + solve(b) << endl;

    return 0;
}