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

3 1 5 1 1 10 2 1 100 3

 

Sample Output

10 48 2302

 

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
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;ir)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;
}