标签:hoj2276 soj2498 count prime 区间素数筛 数论
Description
Given an integer interval [L, R](L <= R <= 2147483647, R - L <= 1000000), please calculate the number of prime(s) in the interval.
InputThere is one line in the input, which contains two integer: L, R.
OutputThere is only one line , which contains the number of prime(s) in the interval.
Sample Input2 11Sample Output
5
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int MAXN = 50000;
const int MAXM = 1e6 + 10;
typedef long long ll;
int isprime[MAXN]; //小数字素数判定
int isprimeB[MAXM]; //大数素数判定
ll prime[MAXN]; //小素数
ll primeB[MAXM]; //大素数
int cnt,cntB;
void getP() //线性素数筛,比普通筛更快
{
for(int i = 1; i < MAXN; i++)
isprime[i] = 1;
for(int i = 2; i < MAXN; i++)
{
if(isprime[i])
{
prime[cnt++] = i;
}
for(int j = 0; j < cnt&&prime[j] * i < MAXN; j++) //采用最小质因子和剩余质因子积的策略来筛
{
isprime[prime[j] * i] = 0;
if(i%prime[j] == 0) break;
}
}
}
void getIntP(ll a,ll b) //区间素数筛
{
for(int i = 0; i <= b - a; i++) isprimeB[i] = 1;
int k;
for(int i = 0; i < cnt&&prime[i]*prime[i]<=b; i++)
{
k = a / prime[i];
if(k*prime[i] < a) k++; //求>=a的第一个素数
if(k <= 1) k++; //如果k==1则表示是第一个质数,当然不能筛掉,事实上k不可能等于0
while(k*prime[i] <= b)
{
isprimeB[k*prime[i] - a] = 0; //筛着走
k++;
}
}
cntB = 0;
for(int i = 0; i <= b - a; i++)
{
if(isprimeB[i])
primeB[cntB++] = i + a;
}
}
int main()
{
ll a, b;
getP();
while(cin >> a >> b)
{
ll ans = 0;
if(b < MAXN)
{
for(int i = a; i <= b; i++)
if(isprime[i]) ans++;
}
else
{
cntB = 0;
getIntP( a, b );
for(int i = 0; i <= b - a;i++)
if(isprimeB[i])ans++;
}
cout << ans << endl;
}
return 0;
}解题报告 之 HOJ2276 SOJ2498 Count prime
标签:hoj2276 soj2498 count prime 区间素数筛 数论
原文地址:http://blog.csdn.net/maxichu/article/details/45607699