POJ－2689－Prime Distance

题目链接

题解

代码

#include <iostream>

using namespace std;

const int MAXN = 100010;
int prime[MAXN + 1];

//  预处理
void getPrime()
{
memset(prime, 0, sizeof(prime));
for (int i = 2; i <= MAXN; i++)
{
if (!prime[i])
{
prime[++prime[0]] = i;
}
for (int j = 1; j <= prime[0] && prime[j] <= MAXN / i; j++)
{
prime[prime[j] * i] = 1;
if (i % prime[j] == 0)
{
break;
}
}
}
}

bool notPrime[1000010];
int primes[1000010];

void getPrimes(int L, int R)
{
memset(notPrime, false, sizeof(notPrime));
if (L < 2)
{
L = 2;
}
for (int i = 1; i <= prime[0] && (long long)prime[i] * prime[i] <= R; i++)
{
int s = L / prime[i] + (L % prime[i] > 0);
if (s == 1)
{
s = 2;
}
for (int j = s; (long long)j * prime[i] <= R; j++)
{
if ((long long)j * prime[i] >= L)
{
notPrime[j * prime[i] - L] = true;
}
}
}
primes[0] = 0;
for (int i = 0; i <= R - L; i++)
{
if (!notPrime[i])
{
primes[++primes[0]] = i + L;
}
}
}

int main(int argc, const char * argv[])
{
//  预处理
getPrime();
int L, U;
while (cin >> L >> U)
{
getPrimes(L, U);
if (primes[0] < 2)    //  素数数量不够成对
{
cout << "There are no adjacent primes.\n";
}
else
{
int xOne = 0, xTwo = 100000000, yOne = 0, yTwo = 0;
for (int i = 1; i < primes[0]; i++)
{
if (primes[i + 1] - primes[i] < xTwo - xOne)    //  排查最近的素数对
{
xOne = primes[i];
xTwo = primes[i + 1];
}
if (primes[i + 1] - primes[i] > yTwo - yOne)    //  排查最远的素数对
{
yOne = primes[i];
yTwo = primes[i + 1];
}
}
cout << xOne << ',' << xTwo << "are closest," << yOne << ',' << yTwo << "are most distant.\n";
}
}

return 0;
}