HDU-6069 Counting Divisors

$Description$
图片说明
$Solution$
区间筛做法.
将 $\sqrt r$ 内的素数筛出来,以素数 $p\cdot k \geq l$ 的位置开始往后筛，边记录每个素数对该数的贡献.
$Code$
#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof(a))
#define sc scanf
#define itn int
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
#define STR clock_t startTime = clock();
#define END clock_t endTime = clock();cout << double(endTime - startTime) / CLOCKS_PER_SEC *1000<< "ms" << endl;
using namespace std;
const int N=1e6+5;
const int mod=998244353;
const int oo=0x7fffffff;
const int sup=0x80000000;
typedef long long ll;
typedef unsigned long long ull;
template <typename it>void db(it *begin,it *end){while(begin!=end)cout<<(*begin++)<<" ";puts("");}
template <typename it>
string to_str(it n){string s="";while(n)s+=n%10+'0',n/=10;reverse(s.begin(),s.end());return s;}
template <typename it>int o(it a){cout<<a<<endl;return 0;}
inline ll mul_64(ll x,ll y,ll c){return (x*y-(ll)((long double)x/c*y)*c+c)%c;}
ll ksm(ll a,ll b,ll c){ll ans=1;for(;b;b>>=1,a=a*a%c)if(b&1)ans=ans*a%c;return ans;}
void exgcd(ll a,ll b,ll &x,ll &y){if(!b)x=1,y=0;else exgcd(b,a%b,y,x),y-=(a/b)*x;}
int prime[N],tot=0;
bool vis[N];
ll cnt[N];
ll l,r,k;
ll a[N];
void pre(){
    for(int i=2;i<N;i++){
        if(!vis[i])prime[++tot]=i;
        for(int j=1;j<=tot&&i*prime[j]<N;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0)break;
        }
    }
}
ll cal(int x,int p){
    ll res=0;
    int ct=0;
    while(a[x]%p==0)a[x]/=p,ct++;
    res=k*ct+1;
    return res%mod;
}
int main(){
   // IN STR
    pre();
    int t;
    cin>>t;
    while(t--){
        sc("%lld%lld%lld",&l,&r,&k);
        for(int i=0;i<r-l+1;i++)cnt[i]=1,a[i]=l+i;
        ll ans=0;
        for(int i=1;i<=tot&&1LL*prime[i]<=r;i++){
            ll x=l/prime[i];
            if(l%prime[i]!=0)x++;
            x=x*prime[i];
            for(ll j=x;j<=r;j+=prime[i]){
                cnt[j-l]*=cal(j-l,prime[i]);
                if(cnt[j-l]>=mod)cnt[j-l]-=cnt[j-l]/mod*mod;
            }
        }
        for(int i=0;i<r-l+1;i++){
            if(a[i]>1)cnt[i]=cnt[i]*(1+k)%mod;
            ans+=cnt[i];
            if(ans>=mod)ans-=ans/mod*mod;
        }
        printf("%lld\n",ans);
    } //END
}