[数论] HOJ 2276 Count prime 筛素数

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Count prime

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Source : SCU Programming Contest 2006 Final Time limit : 1 sec Memory limit : 64 M

Submitted : 732, Accepted : 172

Given an integer interval [L, R](L <= R <= 2147483647, R - L <= 1000000), please calculate the number of prime(s) in the interval.

Input

Multiple cases.
There is one line in each case, which contains two integer: L, R.

Output

There is only one line for each case, which contains the number of prime(s) in the interval.

Sample Input

2 11

Sample Output

解题报告：

看代码，代码如下：

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#define N 500000using namespace std;long long prim[500010];long long visited[500010];long long ranks[500010];long long flag[1000100];void isprime(){    long long num = 0;    memset(visited,0,sizeof(visited));    memset(ranks,0,sizeof(ranks));//求出[1,N]范围内的质数，这些质数将有可能成为[1,2147483647]范围内的质因子 //用线性筛法筛选素数    for(long long i=2; i<=N; i++){        if(visited[i] == 0)            prim[num++] = i;        for(long long j=0; j<num && prim[j]*i<=N; j++){            visited[prim[j]*i] = 1;            if(i%prim[j] == 0)                break;        }        ranks[i] = num;    }}int main(){    long long l,r;    isprime();    while(scanf("%lld%lld",&l,&r)==2){        memset(flag,0,sizeof(flag));        long long ans = 0;        //如果[l,r]在[1,N]范围内        if(r<=N){            printf("%lld\n",ranks[r] - ranks[l-1]);            continue;        }        //计算[l,N]范围内的质数个数        if(l<=N){            ans += ranks[N] - ranks[l-1];            l = N + 1;        }        //计算[l,r]范围内质数个数，l>N>=prim[i]        //方法是排除合数，[1,2147483647]范围内合数的质因子必定在prim[]数组中        //为避免数组越界，减去l做平移        //由于prim[i]*prim[i]可能会超出int精度，都用long long        for(long long i=0; prim[i] * prim[i]<=r; i++){            long long start = 0;            //start为最接近l的，以prim[i]为质因子的合数            if(l%prim[i] == 0)                start = l;            else                start = (l/prim[i]+1)*prim[i];            //将[l,r]范围内的以prim[i]为质因子的合数全部标记            for(; start<=r; start+=prim[i])                flag[start-l] = 1;        }        for(long long i=l; i<=r; i++)            if(flag[i-l] == 0)                ans++;        printf("%lld\n",ans);    }    return 0;}

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