2017杭电多校第四场1003 Counting Divisors (分解质因数） hdu 6069

Counting Divisors

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1374    Accepted Submission(s): 495

Problem Description

In mathematics, the function d(n) denotes the number of divisors of positive integer n.

For example, d(12)=6 because 1,2,3,4,6,12 are all 12's divisors.

In this problem, given l,r and k, your task is to calculate the following thing :

(∑i=lrd(ik))mod998244353

 

Input

The first line of the input contains an integer T(1≤T≤15), denoting the number of test cases.

In each test case, there are 3 integers l,r,k(1≤l≤r≤1012,r−l≤106,1≤k≤107).

 

Output

For each test case, print a single line containing an integer, denoting the answer.

 

Sample Input

31 5 11 10 21 100 3

 

Sample Output

10482302

 

Source

2017 Multi-University Training Contest - Team 4

 

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题意：求那个公式

解题思路：设n=pc11pc22...pcmmn=p，则d(nk)=(kc1+1)(kc2+1)...(kcm+1)枚举不超过r√​​ 的所有质数p，再枚举区间[l,r]中所有p的倍数，将其分解质因数，最后剩下的部分就是超过r√​​ 的质数，只可能是0个或1个。 
时间复杂度O(r√+(r−l+1)loglog(r−l+1))

#include<cstdio>typedef long long ll;const int N=1000010,P=998244353;int Case,i,j,k,p[N/10],tot,g[N],ans;ll n,l,r,f[N];bool v[N];inline void work(ll p){  for(ll i=l/p*p;i<=r;i+=p)if(i>=l){    int o=0;    while(f[i-l]%p==0)f[i-l]/=p,o++;    g[i-l]=1LL*g[i-l]*(o*k+1)%P;//i的因子个数   }}int main(){//枚举到N的所有质数   for(i=2;i<N;i++){    if(!v[i])p[tot++]=i;    for(j=0;j<tot&&i*p[j]<N;j++){      v[i*p[j]]=1;      if(i%p[j]==0)break;    }  }  scanf("%d",&Case);  while(Case--){    scanf("%lld%lld%d",&l,&r,&k);    n=r-l;    for(i=0;i<=n;i++)f[i]=i+l,g[i]=1;    for(i=0;i<tot;i++){      if(1LL*p[i]*p[i]>r)break;      work(p[i]);    }    for(ans=i=0;i<=n;i++){      if(f[i]>1)g[i]=1LL*g[i]*(k+1)%P;       ans=(ans+g[i])%P;    }    printf("%d\n",ans);  }  return 0;}