【线性筛】【质因数分解】【约数个数定理】hdu6069 Counting Divisors

d(x)表示x的约数个数，让你求（l,r<=10^12,r-l<=10^6,k<=10^7）

 

#include<cstdio>
using namespace std;
#define MOD 998244353ll
#define MAXP 1000100
typedef long long ll;
ll x,y;
int T,K;
bool isNotPrime[MAXP+10];
int num_prime,prime[MAXP+10];
void shai()
{for(long i = 2 ; i <  MAXP ; i ++)  {  if(! isNotPrime[i])  prime[num_prime ++]=i;  for(long j = 0 ; j < num_prime && i * prime[j] <  MAXP ; j ++)  {  isNotPrime[i * prime[j]] = 1;  if( !(i % prime[j]))  break;  }  }  
}
ll b[1000010],a[1000010];
int main(){scanf("%d",&T);shai();for(;T;--T){scanf("%lld%lld%d",&x,&y,&K);for(ll i=x;i<=y;++i){a[i-x+1ll]=i;b[i-x+1ll]=1;}for(int i=0;i<num_prime;++i){ll t=x/(ll)prime[i]*(ll)prime[i]+(ll)(x%(ll)prime[i]!=0)*(ll)prime[i];for(ll j=t;j<=y;j+=(ll)prime[i]){int cnt=0;while(a[j-x+1ll]%(ll)prime[i]==0){a[j-x+1ll]/=(ll)prime[i];++cnt;}b[j-x+1ll]=(b[j-x+1ll]*(((ll)cnt*(ll)K%MOD+1ll)%MOD))%MOD;}}ll ans=0;for(ll i=x;i<=y;++i){if((a[i-x+1ll]>1ll)){b[i-x+1ll]=(b[i-x+1ll]*((ll)K+1ll))%MOD;}ans=(ans+b[i-x+1ll])%MOD;}printf("%lld\n",ans);}return 0;
}