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 + ;

 char* isPrime;
 LL bias;

 vector<LL> eratosthenesOddSingleBlock(LL from, LL to)
 {
     vector<LL> primes;
     if (from > to || to <= ) return primes;
     if (to == )
     {
         primes.push_back();
         return primes;
     }
     if (!(from & )) ++from;
     if (from > to) return primes;

     if (from <= ) primes.push_back(), from = ;
     const int memorySize = (to - from) /  + ;
     char* isPrime = new char[memorySize];

     for (int i = ; i < memorySize; i++)
         isPrime[i] = ;
     for (LL i = ; i*i <= to; i+=)
     {
       if (i >= * && i %  == )
          continue;
       if (i >= * && i %  == )
          continue;
       if (i >= * && i %  == )
          continue;
       if (i >= * && i %  == )
          continue;
       if (i >= * && i %  == )
          continue;
       LL minJ = ((from+i-)/i)*i;
       if (minJ < i*i)
           minJ = i*i;
       if ((minJ & ) == )
           minJ += i;
       for (LL j = minJ; j <= to; j += *i)
       {
           int index = j - from;
           isPrime[index/] = ;
       }
     }

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

     delete[] isPrime;
     return primes;
 }

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

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

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

 LL solve(int row)
 {
     if (row == ) return ;
     if (row <= ) return ;

     LL l = getNumber(row - , ), r = getNumber(row + , row + );
     vector<LL> primes = eratosthenesOddSingleBlock(l, r);
     isPrime = new char[r - l + ];
     memset(isPrime, , r - l + );
     bias = l;
     for (auto p: primes) isPrime[p - bias] = ;

     pair<int, int> pos;
     LL res = , val = getNumber(row, );
     for (int col = ; 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 ;
 }