0x31 质数

poj2689 算根号R的质数，然后把L~R区间（这个很小啊）的合数判下

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;

int pr,prime[1100000];bool v[1100000];
void get_prime(int li)
{
    pr=0;memset(v,false,sizeof(v));
    for(int i=2;i<=li;i++)
    {
        if(v[i]==false)prime[++pr]=i;
        for(int j=1;j<=pr&&i*prime[j]<=li;j++)
        {
            v[i*prime[j]]=true;
            if(i%prime[j]==0)break;
        }
    }
}

int blen,bprime[1100000];
int main()
{
    int L,R;
    while(scanf("%d%d",&L,&R)!=EOF)
    {
        if(L==1)L++;
        get_prime( (int(sqrt(double(R)))) );
        memset(v,false,sizeof(v));
        for(int j=1;j<=pr;j++)
        {
            LL d=L%prime[j]==0?L:((LL)prime[j])*((LL)(L/prime[j]+1));
            if(d==prime[j])d+=prime[j];
            while(d<=R)
            {
                v[d-L+1]=true;
                d+=prime[j];
            }
        }
        blen=0;
        for(int i=1;i<=R-L+1;i++)
            if(v[i]==false)bprime[++blen]=i;
        if(blen<=1)printf("There are no adjacent primes.\n");
        else
        {
            int mx1=-1,mx2,mn1=-1,mn2;
            for(int i=1;i<blen;i++)
            {
                if(mx1==-1||mx2-mx1<bprime[i+1]-bprime[i])
                    mx1=bprime[i], mx2=bprime[i+1];
                    
                if(mn1==-1||mn2-mn1>bprime[i+1]-bprime[i])
                    mn1=bprime[i], mn2=bprime[i+1];
            }
            mx1+=L-1, mx2+=L-1, mn1+=L-1, mn2+=L-1;
            printf("%d,%d are closest, %d,%d are most distant.\n",mn1,mn2,mx1,mx2);
        }
    }
    return 0;
}

阶乘分解 筛素数然后看每个素数的贡献啊。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;

int pr,prime[1100000];bool v[1100000];
void get_prime(int li)
{
    pr=0;memset(v,false,sizeof(v));
    for(int i=2;i<=li;i++)
    {
        if(v[i]==false)prime[++pr]=i;
        for(int j=1;j<=pr&&i*prime[j]<=li;j++)
        {
            v[i*prime[j]]=true;
            if(i%prime[j]==0)break;
        }
    }
}

int main()
{
    int n;
    scanf("%d",&n);
    get_prime(n);
    
    for(int j=1;j<=pr;j++)
    {
        printf("%d ",prime[j]);
        LL k=prime[j];int c=0;
        while(k<=n)
        {
            c+=n/k;
            k*=prime[j];
        }
        printf("%d\n",c);
    }
    return 0;
}