HDU 6069 Counting Divisors(枚举区间)(素数筛模版)

本题两个关键点
1.要得知一个数有多少因子,假设他可以被分解为素因数
n = p1^c1*p2^c2+p3^c3…
那么他的因子数为(c1+1)(c2+1)+…+(ck+1).
好像他们又说这是个小学生都知道的结论T^T
2.假如我想要知道[l,r]区间内所有的因子数,不能像单一的一样单个扫描2,3,5,7…而是要2,扫l~r所有2的倍数,3扫,5扫,这样就不会浪费时间在扫一些大素数上.
最后的最后,分解时要忘记本身是大素数的情况!!!!

/*  xzppp  */
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define FFF freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define MP make_pair
#define PB push_back
#define _ %MOD
typedef long long  LL;
typedef unsigned long long ULL;
typedef pair<int,int > pii;
typedef pair pll;
typedef pair<double,double > pdd;
typedef pair<double,int > pdi;
const int MAXN = 1e6+17;
const int MAXM = 1e6+17;
const int MAXV = 2*1e3+17;
const int BIT = 15+3;
const int INF = 0x7fffffff;
const LL INFF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 998244353;
int prime[1100000],primesize=0;
bool isprime[11000000];
void sieve(int listsize){
    memset(isprime, 1, sizeof(isprime));
    isprime[1] = false;
    for(int i=2;i<=listsize;i++){
        if(isprime[i]) prime[++primesize]=i;
         for(int j = 1; j <= primesize && i*prime[j] <= listsize;j++){
            isprime[i*prime[j]] = false;
            if(i%prime[j] == 0) break;
         }
    }
}
LL sum[MAXN],rem[MAXN];
int main()
{
    #ifndef ONLINE_JUDGE 
    FFF
    #endif
    sieve(1000000);
    int t;
    cin>>t;
    while(t--)
    {
        LL l,r,k,ans = 0;
        scanf("%lld%lld%lld",&l,&r,&k);
        for (int i = 0; i < r-l+1; ++i)
        {
            sum[i] = 1;
            rem[i] = i+l;
        }
        for (int i = 1; i <= primesize; ++i)
        {
            LL p = prime[i];
            if(p>r) break;
            LL beg = l/p *p ;
            for (LL j = beg; j <= r; j+=p)
            {
                if(jcontinue;
                LL cnt = 0;
                while(rem[j-l]%p==0)
                {
                    cnt++;
                    rem[j-l]/=p;
                }
                sum[j-l] = (sum[j-l] * (k*cnt+1)_)_;
            }

        }
        for (int i = 0; i < r-l+1; ++i)
        {
            if(rem[i]!=1) sum[i] = (sum[i]*(k+1)_)_;
            ans = (ans + sum[i])_;
        }
        printf("%lld\n",ans );
       }
    return 0;
}