【NOIP模拟题】【暴力求解法】2016.11.17 第一题 素数密度 题解

素数密度

我们发现l，r的距离很小，所以直接暴力。
P.S. 不能把i

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define clr(a,x)  memset(x,a,sizeof(x))
#define ll long long
#ifdef WIN32
#define AUTO "%I64d"
#else
#define AUTO "%lld"
#endif

using namespace std;

const int maxn = 1000005;
const int maxx = 65536;
bool a[maxn],mark[maxn*5];
ll r,l;

template <class T> inline void read(T &xx)
{
    xx = 0;
    T flag = 1;
    char ch = (char)getchar();
    while(ch<'0' || ch>'9')
    {
        if(ch == '-') flag = -1;
        ch = (char)getchar();
    }
    while(ch>='0' && ch<='9')
    {
        xx = (xx<<1) + (xx<<3) + ch - '0';
        ch = (char)getchar();
    }
    xx *= flag;
}

int main(){
    freopen("prime.in","r",stdin);
    freopen("prime.out","w",stdout);
    read(l); read(r);
    clr(true,a); clr(true,mark);
    int ans = 0;
    a[1] = 0;
    for(int i = 2; i <= sqrt(maxx); i++)
        for(int j = 2; j <= maxx/i; j++)
            if(a[i]) a[i*j] = 0;
    if (r <= maxx)
    {
        for(int i = l; i <= r; i++)
            if(a[i] != 0) ans++;
        printf("%d",ans);
        return 0;
    }
    for(int i = 1; i <= maxx; i++)
        if(a[i] != 0)
            for(int j = l/i; j <= r/i; j++)
                mark[i*j-l+1] = 0;
    for(int i = 1; i <= r-l+1; i++)
        if(mark[i]) ans++;
    printf("%d",ans);
    return 0;
}