[欧拉筛] Jzoj P5835 Prime

Description

 

Input

Output

 

Sample Input

Input 1
2 16 2
Input 2
100 1000 2333333
Input 3
10000000000 10000001000 423

Sample Output

Output 1
3
Output 2
561
Output 3
170

 

Data Constraint

 

Hint

 

 

题解

• 对于[1, min(k, √R)]中的每个质数p，将其在[max(L, 2 ∗ p), R]内的倍数标记一下
• 最后没标记的数就是类质数

代码

#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
long long l,r,k,x,y,ans;
bool f[10000010],boo[10000010];
int main()
{
    freopen("prime.in","r",stdin);
    freopen("prime.out","w",stdout);
    scanf("%lld%lld%lld",&l,&r,&k);
    if (k>sqrt(r)) k=sqrt(r);
    memset(f,0,sizeof(f));
    memset(boo,0,sizeof(boo));
    f[1]=1;
    for (long long i=2;i<=k;i++) if (!f[i]) for (long long j=i;j<=k/i;j++) f[i*j]=1;
    for (long long i=1;i<=k;i++)
        if (!f[i])
        {
            if (l<=i&&i<=r) ans^=i;
            x=l/i,y=r/i,x=x*i<l?x+1:x;
            for (long long j=x;j<=y;j++) if (i*j>=l) boo[i*j-l]=1;
        }
    for (long long i=l;i<=r;i++) if (!boo[i-l]) ans^=i;
    printf("%lld",ans);
    return 0;
}